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How to Solve Square Root Problems

Last Updated: August 20, 2024 Fact Checked

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 11 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 406,566 times.

While the intimidating sight of a square root symbol may make the mathematically-challenged cringe, square root problems are not as hard to solve as they may first seem. Simple square root problems can often be solved as easily as basic multiplication and division problems. More complex square root problems, on the other hand, can require some work, but with the right approach, even these can be easy. Start practicing square root problems today to learn this radical new math skill!

Squares & Square Roots Basics

• Try squaring a few more numbers on your own to test this concept out. Remember, squaring a number is just multiplying it by itself. You can even do this for negative numbers. If you do, the answer will always be positive. For example, (-8) 2 = -8 × -8 = 64 .

• As another example, let's find the square root of 25 (√(25)). This means we want to find the number that squares to make 25. Since 5 2 = 5 × 5 = 25, we can say that √(25) = 5 .
• You can also think of this as "undoing" a square. For example, if we want to find √(64), the square root of 64, let's start by thinking of 64 as 8 2 . Since a square root symbol basically "cancels out" a square, we can say that √(64) = √(8 2 ) = 8 .

• On the other hand, numbers that don't give whole numbers when you take their square roots are called imperfect squares . When you take one of these numbers' square roots, you usually get a decimal or fraction. Sometimes, the decimals involved can be quite messy. For instance, √(13) = 3.605551275464...

• 1 2 = 1 × 1 = 1
• 2 2 = 2 × 2 = 4
• 3 2 = 3 × 3 = 9
• 4 2 = 4 × 4 = 16
• 5 2 = 5 × 5 = 25
• 6 2 = 6 × 6 = 36
• 7 2 = 7 × 7 = 49
• 8 2 = 8 × 8 = 64
• 9 2 = 9 × 9 = 81
• 10 2 = 10 × 10 = 100
• 11 2 = 11 × 11 = 121
• 12 2 = 12 × 12 = 144

• Let's say that we want to find the square root of 900. At first glance, this looks very difficult! However, it's not hard if we separate 900 into its factors. Factors are the numbers that can multiply together to make another number. For instance, since you can make 6 by multiplying 1 × 6 and 2 × 3, the factors of 6 are 1, 2, 3, and 6.
• Instead of working with the number 900, which is somewhat awkward, let's instead write 900 as 9 × 100. Now, since 9, which is a perfect square, is separated from 100, we can take its square root on its own. √(9 × 100) = √(9) × √(100) = 3 × √(100). In other words, √(900) = 3√(100) .
• We can even simplify this two steps further by dividing 100 into the factors 25 and 4. √(100) = √(25 × 4) = √(25) × √(4) = 5 × 2 = 10. So, we can say that √(900) = 3(10) = 30 .

• Note that although imaginary numbers can't be represented with ordinary digits, they can still be treated like ordinary numbers in many ways. For instance, the square roots of negative numbers can be squared to give those negative numbers, just like any other square root. For example, i 2 = -1

Using Long Division-Style Algorithms

• Start by writing out your square root problem in the same from as a long division problem. For example, let's say that we want to find the square root of 6.45, which is definitely not a convenient perfect square. First, we'd write an ordinary radical symbol (√), then we'd write our number underneath it. Next, we'd make a line above our number so that it's in a little "box" — just like in long division. When we're done, we should have a long-tailed "√" symbol with 6.45 written under it.
• We'll be writing numbers above our problem, so be sure to leave space.

• In our example, we would divide 6.45 into pairs like this: 6-.45-00 . Note that there is a "leftover" digit on the left — this is OK.

• In our example, the first group in 6-.45-00 is 6. The biggest number that is less than or equal to 6 when squared is 2 — 2 2 = 4. Write a "2" above the 6 under the radical.

• In our example, we would start by taking the double of 2, the first digit of our answer. 2 × 2 = 4. Next, we would subtract 4 from 6 (our first "group"), getting 2 as our answer. Next, we would drop down the next group (45) to get 245. Finally, we would write 4 once more to the left, leaving a small space to add onto the end, like this: 4_.

• In our example, we want to find the number to fill in the blank in 4_ × _ that makes the answer as large as possible but still less than or equal to 245. In this case, the answer is 5 . 45 × 5 = 225, while 46 × 6 = 276.

• Continuing from our example, we would subtract 225 from 245 to get 20. Next, we would drop down the next pair of digits, 00, to make 2000. Doubling the numbers above the radical sign, we get 25 × 2 = 50. Solving for the blank in 50_ × _ =/< 2,000, we get 3 . At this point, we have "253" above the radical sign — repeating this process once again, we get a 9 as our next digit.

• In our example, the number under the radical sign is 6.45, so we would simply slide the point up and place it between the 2 and 5 digits of our answer, giving us 2.539 .

Quickly Estimating Imperfect Squares

• For example, let's say we need to find the square root of 40. Since we've memorized our perfect squares, we can say that 40 is in between 6 2 and 7 2 , or 36 and 49. Since 40 is greater than 6 2 , its square root will be greater than 6, and since it is less than 7 2 , its square root will be less than 7. 40 is a little closer to 36 than it is to 49, so the answer will probably be a little closer to 6. In the next few steps, we'll narrow our answer down.

• In our example problem, a reasonable estimate for the square root of 40 might be 6.4 , since we know from above that the answer is probably a little closer to 6 than it is to 7.

• Multiply 6.4 by itself to get 6.4 × 6.4 = 40.96 , which is slightly higher than original number.
• Next, since we over-shot our answer, we'll multiply the number one tenth less than our estimate above by itself and to get 6.3 × 6.3 = 39.69 . This is slightly lower than our original number. This means that the square root of 40 is somewhere between 6.3 and 6.4 . Additionally, since 39.69 is closer to 40 than 40.96, you know the square root will be closer to 6.3 than 6.4.

• In our example, let's pick 6.33 for our two-decimal point estimate. Multiply 6.33 by itself to get 6.33 × 6.33 = 40.0689. Since this is slightly above our original number, we'll try a slightly lower number, like 6.32. 6.32 × 6.32 = 39.9424. This is slightly below our original number, so we know that the exact square root is between 6.33 and 6.32 . If we wanted to continue, we would keep using this same approach to get an answer that's continually more and more accurate.

Calculator, Practice Problems, and Answers

Expert Q&A

• For quick solutions, use a calculator. Most modern calculators can instantly find square roots. Usually, all you need to do is to simply type in your number, then press the button with the square root symbol. To find the square root of 841, for example, you might press: 8, 4, 1, (√) and get an answer of 29 . [16] X Research source Thanks Helpful 0 Not Helpful 0

You Might Also Like

• ↑ David Jia. Academic Tutor. Expert Interview. 14 January 2021.
• ↑ https://virtualnerd.com/algebra-foundations/powers-square-roots/powers-exponents/squaring-a-number
• ↑ https://www.mathsisfun.com/square-root.html
• ↑ http://virtualnerd.com/algebra-1/algebra-foundations/powers-square-roots/square-roots/square-root-estimation
• ↑ https://www.cuemath.com/algebra/perfect-squares/
• ↑ https://www.khanacademy.org/math/algebra/rational-exponents-and-radicals/alg1-simplify-square-roots/a/simplifying-square-roots-review
• ↑ https://math.libretexts.org/Courses/Monroe_Community_College/MTH_165_College_Algebra_MTH_175_Precalculus/00%3A_Preliminary_Topics_for_College_Algebra/0.03%3A_Review_-_Radicals_(Square_Roots)
• ↑ http://www.homeschoolmath.net/teaching/square-root-algorithm.php
• ↑ https://www.cuemath.com/algebra/square-root-by-long-division-method/
• ↑ https://virtualnerd.com/algebra-1/algebra-foundations/powers-square-roots/square-roots/square-root-estimation
• ↑ http://www.math.com/students/calculators/source/square-root.htm

About This Article

To solve square root problems, understand that you are finding the number that, when multiplied by itself, equals the number in the square root. For quick recall, memorize the first 10-12 perfect squares, so that you recognize the square root of numbers like 9, 25, 49, or 121. If possible, break the number under the square root into individual perfect squares. For example, √(900) can be broken into √(9) × √(100), and √(100) can be broken into √(25) × √(4), reducing the problem to √(9) × √(25) × √(4), or 3 x 5 x 2 for an answer of 30. If you want to learn how to estimate imperfect square roots, keep reading the article! Did this summary help you? Yes No

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Square root

Square roots

Here is everything you need to know about square numbers and square roots. You will learn what a square number is and how to find the square roots of numbers and expressions. You will also apply this knowledge to problem solve.

Students first work with square roots in 8 th grade math and expand that knowledge as they move into high school math classes.

What is a square number?

A square number is a number that is the result of multiplying that number to itself, or “squared”. In other words, a square number is found when multiplying any number by itself. The solution when a number is multiplied to itself is called a perfect square.

Look at this table to see the relationship.

Square numbers can also be represented by an array that forms squares.

You can arrange 1^2 as a square which has side lengths 1 unit.

You can arrange 2^2 as a square which has side lengths 2 units.

You can arrange 3^2 as a square which has side lengths 3 units.

You can arrange 4^2 as a square which has side lengths 4 units.

There are an infinite number of perfect square numbers, which makes it impossible to know all of them. However, remembering the perfect squares up to 15 \times 15 is extremely helpful for problem solving purposes.

[FREE] Square Roots Worksheet (Grade 8)

Use this worksheet to check your 8th grade students’ understanding of square roots. 15 questions with answers to identify areas of strength and support!

Squaring negative numbers

Negative numbers can be squared like positive numbers.

For example:

Notice how (- \; 5)^2=(5)^2=25

The square of any negative number is always positive as you are multiplying a negative by a negative, which gives you a positive answer.

What is a square root?

The square root is a factor of a number that, when multiplied to itself, gives the original number. In other words, square rooting a number is the inverse operation of squaring a number. Taking a square root undoes squaring a number. The square root symbol (square root function) looks like this:

Its mathematical name is a radical or radical symbol.

The square root of a number is represented by:

\sqrt{4} where 4 is considered the radicand because it is the number under the radical symbol.

\sqrt{4}=2. In this case, 2 is the principal square root because it is the positive square root.

Since 2\times{2}=4, and (- \; 2)\times(- \; 2)=4, then \sqrt{4}=2 or - \; 2. This is sometimes written as \sqrt{4}=\pm \; {2}.

Algebraic expressions such as variables can be squared. For example, x^2 means x\times{x}. So, \sqrt{x^2}=x. Taking the square root undoes the squaring.

x^4 is also a square term because x^{2}\times{x}^{2}=x^{4} so, \sqrt{x^{4}}=x^{2}

You can simplify square root expressions by breaking the expression down into perfect square factors.

Notice how \sqrt{32} is not a perfect square. The expression can still be simplified by breaking the number into perfect square factors. Specifically, look for the largest square factor of the number under the radical.

In this case, the largest square factor of 32 is 16. So, \sqrt{32} can be rewritten as \sqrt{16\times{2}} which is \sqrt{16}\times\sqrt{2}.

Since 16 is a perfect square, you can take the square root, but \sqrt{2} is not a perfect square, so you cannot take the square root without a calculator; it’s considered to be an irrational number.

\sqrt{32} fully simplified is 4\sqrt{2}.

See also : How to simplify radicals

Common Core State Standards

How does this relate to 8 th grade math?

• Grade 8: Expressions and Equations (8.EE.A.2) Use square root and cube root symbols to represent solutions to equations of the form x^{2}=p and x^{3}=p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that \sqrt{2} is irrational.

How to find square roots

In order to find square roots:

Write an expression using the \bf{\sqrt{\;}} (radical sign).

Find the square root, simplify if necessary.

Write the simplified answer.

Square numbers and square roots examples

Example 1: square root of a perfect square.

Find the square root of 81.

The square root of 81 is expressed as \sqrt{81} .

2 Find the square root, simplify if necessary.

\sqrt{81} is a whole number because 81 is a perfect square.

3 Write the simplified answer.

\sqrt{81}=9 and \sqrt{81}=- \; 9 because 9\times{9}=81 and - \; 9\times{- \; 9}=81

Example 2: square root of a fraction

Find the square root of \cfrac{1}{16}.

The square root of \cfrac{1}{16} is represented by \sqrt{\cfrac{1}{16}} .

\sqrt{\cfrac{1}{16}} is a perfect square because 1 and 16 are perfect square numbers. You can rewrite it with the radical in the numerator and the radical in the denominator.

\sqrt{\cfrac{1}{16}}=\cfrac{\sqrt{1}}{\sqrt{16}}

\cfrac{\sqrt{1}}{\sqrt{16}}=\cfrac{1}{4} or - \; \cfrac{1}{4}.

\sqrt{\cfrac{1}{16}}=\cfrac{\sqrt{1}}{\sqrt{16}}=\cfrac{1}{4} or - \; \cfrac{1}{4}.

Example 3: square root of an algebraic expression

Find the square root of 225s^{4}.

The square root of 225s^{4} is represented by \sqrt{225s^{4}}

\sqrt{225s^{4}} is a perfect square expression because 225 is a perfect square number and s^4 is a perfect square expression.

15\times{15}=225

s^{2}\times{s}^{2}=s^{4}

How to find square roots of algebraic expressions with non-perfect squares

In order to find square roots of algebraic expressions with non-perfect squares:

Find the largest square factor(s) of the term under the root.

Rewrite the radical as a product of the square factor(s) and the other factors.

Simplify the radical expression.

Example 4: square root of an algebraic expression

Simplify \sqrt{8x^{3}}.

Square numbers are 1, 4, 9, 16, 25, …

The largest square factor of 8 is 4 because 4\times{2}=8

The largest square factor of x^3 is x^2 because x^{2}\times{x}=x^{3}

Take the square root of 4 and x^2 because both are perfect squares.

How to solve problems involving square numbers and square roots

In order to problem solve with square numbers and square roots:

Identify whether you need to square or square root the number/variable.

Perform the operation.

Clearly state the answer within the context of the question.

Example 5: word problems with square numbers

The area of a square is 36\mathrm{~in}^{2}. Using the formula to find area of a square, \text{Area}=s^{2}, find the side length of the square.

Identify whether you need to square or square root the number/ variable.

The question is to find the side length, so using the formula you are taking the square root.

\text{Area}=s^{2} where the area is 36\mathrm{~in}^{2}.

Since the question is asking for side length, the positive value is the solution not the negative value.

The side length of the square is 6 inches.

Example 6: word problems with square roots

Two times a number squared is 32. Calculate the possible value(s) of the number.

In this case, you will have to take the square root to find the solution.

Two times a number squared is 32 can be written as the equation 2\times{x}^{2}=32.

For this problem, both + \; 4 and - \; 4 are the solutions.

Check both values in the original equation to validate your answer.

2\times(4)^{2}=32

2\times(- \; 4)^{2}=32

Teaching tips for square numbers and square roots

• Incorporate discovery based learning activities so that students can be like mathematicians and explore perfect square numbers and square roots on their own.
• If students struggle with multiplication tables, have them use a digital, online square root calculator.

Easy mistakes to make

• Squaring numbers incorrectly For example, thinking that squaring a number means to double it. 3^{2} ≠ {6} 3^{2}=3\times{3}=9
• Forgetting about the negative root when taking a square root For example, only recognizing positive values of square roots of integers. \sqrt{100}=10 (only thinking the answer is 10 ) \sqrt{100}=\pm \; {10}

Related laws of exponents lessons

• Laws of exponents
• Exponential notation
• Anything to the power of 0
• Exponent rules
• Negative exponents
• Multiplying exponents
• Dividing exponents
• Distributing exponents

Practice square numbers and square roots questions

1. What is the square root of the number, 169?

Taking the square root of 169 means \sqrt{169}. 169 is a perfect square number because 13\times13=169 and – \; 13\times{- \; 13}=169

So, \sqrt{169}=\pm \; {13}

2. What is the square root of a^{6}?

The square root of a^6 is represented by \sqrt{a^6}. a^6 is a perfect square term because a^{3}\times{a^{3}}=a^{6} so, \sqrt{a^{6}}=a^{3}

3. What are the values of \sqrt{144}?

12 and – \; 12

72 and – \; 72

12 and – \; 11

\sqrt{\;} means square root, so, \sqrt{144} means to take the square root of 144. In other words, what number multiplied to itself is 144?

12\times{12}=144 and – \; 12\times{- \; 12}=144.

4. Simplify the square root expression, \sqrt{36 y^{3}}.

Look for the perfect square factors of 36y^3 in order to simplify it.

36 is a perfect square and y^{3}=y^{2}\times{y} where y^{2} is a perfect square term.

Take the square root of the perfect square terms

\sqrt{36y^{3}} simplified is 6y\sqrt{y}

5. Find the side length of a square with an area of 81\mathrm{~cm}^{2}.

A square has four equal sides and to find the area of a square you take the side length and square it. So, if you have the area and need to find the side length, take the square root.

Since this is the length of the side of a square only the positive value works. So the side length of the square is 9 \, cm.

6. 3 times a number squared is 363. Find the number(s).

3 times a number squared is 363 translates to be:

In this case, both + \; 11 and – \; 11 work. Check both solutions to verify.

Square numbers and square roots FAQs

If you take the square root of a negative number on a calculator, you will get “error.” The reason being that the square root of a negative number is not a real number, it is an imaginary number which is part of the complex number system. Typically in 8 th grade math, you will be asked to find the square root of natural numbers, positive whole numbers, positive integers, and positive real numbers (decimals and fractions).

In algebra 1 and algebra 2 classes you will learn how to find the square root of polynomials. There are polynomials that are considered to be perfect square expressions such as perfect square trinomials.

To solve for the missing side of right triangles you will use the Pythagorean Theorem (Pythagoras’ Theorem), a^2+b^2=c^2. Since the theorem has square terms, you will have to use square root to find missing side lengths. See also : Pythagorean Theorem

Quadratic equations have a squared term and square root undoes squaring. So, you will need to sometimes use square root to solve quadratic equations.

Yes, you can take the square root of prime numbers but they will be considered irrational numbers. So, either leave them under the radial symbol such as \sqrt{5} or use a square root calculator to find the answer. Remember the answer in a calculator will have many decimal places because irrational numbers are non-repeating, non-terminating numbers, so you will have to round the answer.

The next lessons are

• Scientific notation
• Quadratic graphs

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Solving Radical Equations

How to solve equations with square roots, cube roots, etc.

Radical Equations

We can get rid of a square root by squaring (or cube roots by cubing, etc).

Warning: this can sometimes create "solutions" which don't actually work when we put them into the original equation. So we need to Check!

Follow these steps:

• isolate the square root on one side of the equation
• square both sides of the equation

Then continue with our solution!

Example: solve √(2x+9) − 5 = 0

Now it should be easier to solve!

Check: √(2·8+9) − 5 = √(25) − 5 = 5 − 5 = 0

That one worked perfectly.

More Than One Square Root

What if there are two or more square roots? Easy! Just repeat the process for each one.

It will take longer (lots more steps) ... but nothing too hard.

Example: solve √(2x−5) − √(x−1) = 1

We have removed one square root.

Now do the "square root" thing again:

We have now successfully removed both square roots.

Let us continue on with the solution.

It is a Quadratic Equation! So let us put it in standard form.

Using the Quadratic Formula (a=1, b=−14, c=29) gives the solutions:

2.53 and 11.47 (to 2 decimal places)

Let us check the solutions:

There is really only one solution :

Answer: 11.47 (to 2 decimal places)

See? This method can sometimes produce solutions that don't really work!

The root that seemed to work, but wasn't right when we checked it, is called an "Extraneous Root"

So checking is important.

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Radical Expression

10.1  Definitions and Notation

The nth powers of  2,a,3^2 , and  b^3 are, respectively, 2 2^n,a^n,3^(2n) , and b^(3n) . When  n is an even number, the nth power of a positive or a negative number is a positive number. For example,

     (02)^4=16   and   (-2)^4=16

  When n is an odd number, the nth power of a positive number is a positive number, and the nth power of a negative number is a negative number. For example,

     (+3)^3=27   and   (-3)^3=-27 .

DEFINITION   The n th root of a real number a is denoted by root(n,a) . It is a number whose n th power is a ; that is,  (root(n,a))^n with the following conditions:

  1. When  n is an even number and  a>0 , root(n,a)>0 , called the principal root.

    When  n is an even number and   a<0 , root(n,a) is not a real number.

  2. When  n is an odd number and   a>0 , root(n,a)>0 .

    When  n is an odd number and   a<0 , root(n,a)<0

  The n in root(n,a)  (always a natural number greater than 1) is called the index or the order of the radical, and a is called the radicand. When there is no indicated index, as in root(a) , the index  2 is implied and it is read “the square root of a ." When the index is  3 as in root(3,a) , it is read “the cube root of a ."

EXAMPLES   1.  root(49) = root(7^2) = (root(7))^2 = 7

        2.  -root(25)=-root((5)^2)=-(root(5))^2=-5

        3.  root(-4) is not a real number

        4.  root(3,8) = root(3,(2)^3) = (root(2))^3 = 2

        5.  root(5,-32) = root(5,(-2)^5 = (root(5,-2)^5 = -2 = -root(5,32)

        6.  root(4,x^8 = root(4,(x^2)^4) = (root(4,2))^4 = x^2

        7.  root(5,x^15) = root(5,(x^3)^5 = (root(5,x^3))^5 = x^3

  From the definition of fractional exponents (page 321) and the definition of radicals, for  a {is-in} {real} ,  m,n {is-in} {natural} , we have

     root(n,a)=a^(1/n)   and   root(n,a^m)=a^(m/n)

  providing  root(n,a) and  a^(1/n) are defined.

  The above relations enable us to express radicals as fractional exponents and fractional exponents as radicals.

EXAMPLES   1.  root(5,3)=3^(1/5)

        2.  root(3,2^2)=2^(2/3)

        3.  root(x+3)=(x+3)^(1/2)

        4.  x^(4/5)=root(5,x^4)

        5.  3x^(3/4)=3root(4,x^3)

        6.  x^(1/2)y^(2/3)=root(x)root(3,y^2)

  When the value of a radical expression is a rational number, we say it is a perfect root Since root(n,a^nk)=a^k , a radical expression is a perfect root if the radicand can be expressed as a product of factors each to an exponent that is an integral multiple of the radical index.

  The value of the radical is obtained by forming the product of the factors. where the exponent of each factor is its original exponent divided by the radical index.

EXAMPLES   1.   root(5^6) = 5^(6/2) = 5^3

        2.   root(x^10) = x^(10/2) = x^5

        3.   root(3,8x^6y^9 = root(3,2^3x^6y^9 = 2^(3/3)x^(6/3)y^(9/3) = 2x^2y^3

Note   Nonperfect roots such as  root(2),root(3,2),root(3),root(4,5),root(5,4),1+root(2) and  5-root(3,9) are irrational numbers. An irrational number is a number that cannot be expressed in the form  p/q , where  p,q ∈ {Iota} , q!=0 .

Note   1. Since  a^(m/n)=a^(mk/nk) for all  a>0 ,  a {is-in} {real} , and  m,n ∈ N ,  k {is-in} {rat} ,  k>0 , we have  root(n,m)=root(nk,a^mk) , provided  nk and  mk ∈ N .

     root(3,a)=root(6,a^2)   and   root(n,1)=1

  2.  1^n=1   and   root(n,1)=1 .

10.2  Standard Form of Radicals

THEOREM    If  a,b ∈ R ,  a>0 , b>0 , and  n ∈ N then  root(n,ab)=root(n,a)root(n,b) .

Proof      root(n,ab) = (ab)^(1/n) = a^(1/n)b^(1/n) = root(n,a)root(n,b)

EXAMPLES   1.  root(32) = root(2^5) = root(2^4*2^1)

                  = root(2^4)*root(2)

                  = 2^2root(2) = 4root(2)

        2.  root(16x^3y = root(2^4x^3y = root(2^4x^2xy)

                        = root(2^4x^2)root(xy)

                        = 2^2x root(xy)

                        = 4x root(xy)

Let's see some more problems and our step by step solver will simplify the radical expressions. Please click "Solve Similar" for more examples.

        3.  root(3,27x^2y^4) = root(3,3^3x^2y^4) = root(3,3^3z^2y^3y

                            = root(3,3^3y^3)root(3,x^2y)

                           = 3y root(3,x^2y)

  The expression  3y root(3,x^2y) is called the standard form of the expression root(3,27x^2y^4) .

  A radical expression is said to be in standard form if the following conditions hold:

  1. The radicand is positive.

  2. The radical index is as small as possible.

  3. The exponent of each factor of the radicand is a natural number less than the radical index.

  4. There are no fractions in the radicand.

  5. There are no radicals in the denominator of a fraction.

  By simplifying a radical expression, we mean putting the radical expression in standard form.

  When the radicand is negative, the definition gives us the following:

  When  n is even and  a>0 ,   root(n,-a) is not a real number.

  When  n is odd and  a>0 ,   root(n,-a) = -root(n,a) .

EXAMPLES   1.  root(3,-5)=-root(3,5)

        2.  root(5,-x^2y^3)=-root(5,x^2y^3)

When the radical index and the exponents of all the factors in the radicand have a common factor, divide both the radical index and the exponents of the factors of the radicand by their common factor That is, apply root(nk,a^mk)=root(n,a^m) to obtain the smallest possible radical index.

EXAMPLE     root(6,a^2b^4)=root(3,ab^2)

When the exponents of some factors of the radicand are greater than the radical index, but not an integral multiple of it, write each of these factors as a product of two factors one factor with an exponent that is an integral multiple of the radical index, and the other factor with an exponent that is less than the radical index. For example,

       root(2,x^7)=root(3,x^6*x)

Then apply the theorem root(n,ab)=root(n,a)root(n,b) . Write the factors with exponents that are integral multiples of the index under one radical, thus obtaining a perfect root. and the other factors with exponents less than the radical index under the other radical.

EXAMPLE     root(3,x^7)=root(3,x^6*x)=root(3,x^6)root(3,x)=x^2root(3,x)

The cases when there are fractions in the radicand and radicals in the denominator of a fraction will be discussed later. 

EXAMPLE      Put  root(2^3x^5) in standard form.

Solution    root(2^3x^5) = root((2^2*2)(x^4*x)

              = root(2^2x^4)root(2x)

              = 2x^2root(2x)

EXAMPLE      Put  root(8x^3y^2z^5) in standard form.

Solution    root(8x^3y^2z^5) = root(2^3x^3y^2z^5)

                  = root((2^2*2)(x^2*x)y^2(z^4.z)

                 = root(2^2x^2y^2z^4)root(2xz)

                 = 2xyz^2root(2xz)

EXAMPLE      Put  root(3,2^4x^6y^5z^10)  in standard form.

  Solution    root(3,2^4x^6y^5z^10) = root(3,(2^3*2)x^6(y^3*y^2)(z^9*z)

                 =  root(3,2^3x^6y^3z^9)root(3,2y^2z)

                = 2x^2yz^3root(3,2y^2z)

EXAMPLE      Put  root(3,-2x^11y^4 in standard form.

Solution      root(3,-2x^11y^4 = -root(3,2(x^9*x^2)(y^3*y))

                   = -root(3,(x^9y^3))root(3,2x^2y)

                   = -x^3y root(3,2x^2y)

EXAMPLE      Put  root(4,64x^4y^10) in standard form.

Solution      root(4,64x^4y^10) = root(4,2^6x^4y^10)

                  = root(4,(2^4*2^2)x^4(y^8*y^2))

                  = root(4,2^4x^4y^8)root(4,2^2y^2)

                  = 2xy^2root(2y)

10.3  Combination of Radical Expressions

DEFINITION   Radical expressions are said to be similar when they have the same radical index and the same radicand.

EXAMPLES   1. The redical expressions  3root(2) and  5root(2) are similar.

        2. The redical expressions  root(24) and  root(54) can be shown to be similar.

         root(24) = root(2^3*3) = root(2^2*2*3 = 2root(2)

        and  root(54) = root(2*3^3) = root(2*3^2*3 = 3root(6)

        3. The redical expressions root root(18) and  root(27) are not similar.

         root(18) = root(2*3^2) = 3root(2)

        and  root(27) = root(3^3) = root(3^2*3) = 3root(3) .

Radical expressions can be combined only when they are similar. First we put the radical expressions in standard form and then combine similar radicals using the distributive law.

EXAMPLE   Simplify  root(54)-root(24)+root(150) and combine similar redical expressions.

Solution    root(54)-root(24)+root(150)

         = root(2*3^3)-root(2^3*3)+root(2*3*5^2)

        = 3root(6)-2root(6)+5root(6)

        = (3-2+5)root(6)

        = 6root(6)

EXAMPLE   Simplify  x root(147y^3) + y root(75x^2y) - root(48x^2y^3) and combine similar redicals.

Solution    x root(147y^3) + y root(75x^2y) - root(48x^2y^3)  

       = x root(3*7^2y^3)+y root(3*5^2x^2y)-root(2^4*3x^2y^3)

      = 7xy root(3y)+5xy root(3y)-4xy root(3y)

      = 8xy root(3y)

Let's see some more problems and our step by step solver will simplify the combination of radical expressions. Please click "Solve Similar" for more examples.

EXAMPLE   Simplify  3root(8)-root(3,81)-root(128)+root(3,375) and combine similar redicals .

Solution    3root(8)-root(3,81)-root(128)+root(3,375)

       = 3root(2^3)-root(3,3^4)-root(2^7)root(3,3*5^3)

      = 3*2root(2)-3root(3,3)-2^3root(2)+5root(3,3)

      = 6root(2)-3root(3,3)-8root(2)+5root(3,3)

      = (6-8)root(2)+(-3+5)root(3,3)

      = -2root(2)+2root(3,3)

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Root Calculator

Table of contents

Welcome to the root calculator , where we'll go through the theory and practice of how to calculate the nth root of a number , also called the nth radical , together.

We'll start with a quick explanation of what a root is in math and give some easy examples that you might have already seen, like the square root of 2, square root of 3, or the cube root of 4. But what if it's the fourth root that you'd like to find? The previous ones were fairly simple, but what is, say, the 4th root of 81? No worries, we'll show you soon enough!

Sit back, relax, and enjoy the ride through the world of radicals !

What is a root in math?

We all know multiplication, right? Like 12 × 4 = 48 12 \times 4 = 48 12 × 4 = 48 ? If we want to multiply the same number a few times, then we can write it in a simplified form :

Where the small 5 5 5 is called the exponent and means how many copies of the big number (in this case, 12 12 12 ) we take. We also call this operation taking the 5 5 5 -th power of 12 12 12 . You can explore this mathematical operation at Omni's exponent calculator .

A root is the opposite operation. To connect it to the biological meaning, when we look at a grown tree, we see its leaves and trunk, but it is all built upon its roots . And the story is very similar with numbers: when we see the number 125 125 125 , then taking its root will show us the tiny grain that it grew out of . In this example, it will show us that the seed is 5 5 5 because 5 3 = 125 5^3 = 125 5 3 = 125 .

Formally, the n t h n^{\mathrm{th}} n th root of a number a a a is the number b b b , such that:

For instance, let's take a closer look at what is the square root of some number . Suppose that you're digging a swimming pool in your backyard. You'd like it to be as long as it is wide and, all in all, cover an area of 256 256 256 square feet. How can you figure out how long the sides should be ? That's right – by calculating the radical! In this case, it should be the square root of the area, i.e., the square root of 256 256 256 .

And what is the square root of that number? Well, let's see how we can find it and how to calculate the square root in general.

How to calculate the square root

Sometimes calculating the root in math may resemble a guessing game . But it's not the same as rolling dice with your eyes closed, and guess what you'll get. It's more of a calculated guess . After all, once we know that 3 4 = 81 3^4 = 81 3 4 = 81 , we can safely say that the 4 t h 4^{\mathrm{th}} 4 th root of 81 81 81 is 3 3 3 . But we have to know that first.

So what can we do if we forget our handy table of the first one hundred numbers and their first few powers at home? Is it a lost cause? Fortunately, no. Not entirely, but we'll come back to this in a second.

As an example, we'll show how to calculate the square root of 72 72 72 . Our main tool here will be prime factorization, i.e., splitting 72 72 72 into its smallest pieces possible.

In the prime factorization procedure, we take a number (in our case, 72 72 72 ) and find the smallest prime number that divides it . Recall that a prime number is an integer that has only two divisors: 1 1 1 and itself. It's fairly easy to see that for us, it will be 2 2 2 since:

The next step is to find the smallest prime number of the result of the division of the number by the prime number, i.e., the number 36 36 36 . If we continue this until we reach 1 1 1 , we'll get the following primes: 2 2 2 , 2 2 2 , 2 2 2 , 3 3 3 , 3 3 3 . This is the prime factorization of 72 72 72 , and it means that:

Is something not clear about prime factorization? No worries, it's a pretty interesting mathematical problem, sometimes hard to solve, even for computers! You can learn more (almost everything) about it at Omni's prime factorization calculator .

Now, if we find pairs among the same numbers, we'll see that we have a couple of 2 2 2 's, a couple of 3 3 3 's, and a single 2 2 2 is left. This allows us to write the square radical of 72 72 72 as:

A keen eye will observe that the only numbers that stay under the root are exactly the loners that didn't find a pair .

But what about the 2 2 2 ? What is the square root of 2 2 2 ? Well, that's what the " not entirely " was all about. The square root of 2 2 2 , the square root of 3 3 3 , or any other prime number takes us back to a guessing game. Fortunately, we can use our root calculator to figure out that 2 ≈ 1.4142 \sqrt{2} \approx 1.4142 2 ​ ≈ 1.4142 , which gives us:

In essence, when we're asked " what is the square root of..., ", we should first do prime factorization to break down the problem, and if (as above) we're left with some small digit in the end, we just have to use a tool like the root calculator to find it.

" But what about higher radicals? What if I need, e.g., the fourth root of a number? " Well, how convenient of you to ask! That is precisely the problem we'll deal with in the next section .

🙋 For a more exhaustive description of this operation, visit Omni's square root calculator !

Cube root, fourth root, n-th root

Recall how you wanted to dig a swimming pool in the first section . Now suppose that you'd like the whole thing to be a cube that holds 1 , 728 1,728 1 , 728 cubic feet of water. (Don't ask us why. Perhaps anything above is taxed differently?)

How can you find the side of such a pool? Yup – by calculating the cube root of the number (that's where the name cubic root comes from). It will tell us that the length should be:

But how did we get there? Fortunately, the main tool here is the same: prime factorization . If we apply the procedure to 1728 1728 1728 we'll get that:

Now comes the thing that's different – instead of pairs, we group the numbers into triples . That's what the small 3 3 3 in the root symbol suggests – we need third powers . Note that the square roots are, in fact, radicals of order 2 2 2 , but we don't write the 2 2 2 because... Well, if we don't have to do it from one type of root, it might as well be the simplest one . That's just convention and tradition. Think of it as a mathematical equivalent of roasting a turkey for Thanksgiving.

Anyway, coming back to our problem, the grouping allows us to write:

If we go higher with the order of the radical, the same rule applies . When calculating the fourth root, we group the primes into quadruples . Like if you need the 4 t h 4^{\mathrm{th}} 4 th root of 81 81 81 , you first observe that:

So we have four 3 3 3 's. This means that the 4 t h 4^{\mathrm{th}} 4 th root of 81 81 81 is equal to 3 3 3 . And if we need the nth root , we take groups of n n n elements. And, if anything is left after the factorization, we just find it with some external tool like our root calculator .

Alright, after all this time reading through the theory, it's high time we look at a real-life example, and see the root calculator in action , don't you think?

🙋 As for the square root, we have a tool entirely dedicated to the cube root: the cube root calculator !

Example: using the root calculator

Congratulations, it's a boy! Now that you've become a parent, you decide to start early and save some money for when he goes to college. You choose to take a good slice of your savings and leave it in the bank for the next eighteen years so that the amount grows together with your kid.

Suppose that you've managed to put away a solid $ 8 ,  ⁣ 000 \text\textdollar8,\!000 $ 8 , 000 (call this quantity s t a r t \mathrm{start} start ). Unfortunately, somehow you forgot the interest rate on the investment, but what's done is done. The amount at the end will be as much of a surprise to you as it is to your son .

Time passes, years go by, and finally, it's time to gift your kid the money that you saved . You call the bank, and it turns out that there's $ 12 ,  ⁣ 477.27 \text\textdollar12,\!477.27 $ 12 , 477.27 in the account (we'll call this variable e n d \mathrm{end} end ). Not too bad, is it? It seems like you'll be able to make your son's dreams come true.

But, just for yourself, just out of sheer curiosity, can we calculate the interest rate from the numbers we have?

Sure we can , and the root calculator will help us!

Assume that the interest was added to the account at the end of each year and that the money was not taxed at all (yeah, we know we're going a bit far-fetched right there). Then the amount we get in the end is described by the formula :

Where the 1 8 t h 18^{\mathrm{th}} 1 8 th power comes from the eighteen years that the money spent in the bank: this is the formula you can meet at the simple interest calculator . In our case, this translates to:

If we divide both sides by $ 8 ,  ⁣ 000 \text\textdollar8,\!000 $ 8 , 000 , we'll get that:

Or after an approximation:

So if we have the 18 t h 18\mathrm{th} 18 th power on the right, we need to find the 18 t h 18\mathrm{th} 18 th radical of the number on the left . Now, this is something slightly more complicated than the square root of 3 3 3 , isn't it?

We turn to our root calculator . In there, we have two numbers: a a a and n n n . When we look at the symbolic picture in there, we see that n n n is the order of the root , so we input n = 18 n = 18 n = 18 . In turn, a a a is the number under the radical , so we take a = 1.5597 a = 1.5597 a = 1.5597 . This makes the root calculator spit out the answer to be:

If we translate the decimal into percentages, we obtain:

It seems fairly small, but oh, how it grew in eighteen years!

Alright, curiosity satisfied , time to go back to the birthday cake. Let's just hope that your son will make good use of the money and keep to his studies.

How do I find the square root without a calculator?

One method of estimating square roots is called the Babylonian method . Start with a guess of the square root and divide the original number by your guess. Then, take an average of your guess and the result of the division. Continue until you obtain the required precision.

For example, to find the square root of 2 :

• Start with a guess of 1.5.
• Divide 2 by 1.5, which is 1.3333.
• Find the average of 1.5 and 1.3333, which is 1.4167.
• Repeat steps 2 and 3 until the result settles down to the required accuracy.

How do I find the cube root on a calculator?

To find the cube root of a number on a regular calculator, enter the number and then press the ∛x button . On a scientific graphing calculator, press the ∛ button and then enter the number.

To find higher roots on a regular calculator, enter the order of the root, press the y √x button, and then finally the number. The steps are the same on a scientific calculator, though the button might be labeled x √.

How do I calculate a square root in Excel?

To calculate the square root of a number in Excel or Google Sheets, use the SQRT() function . For example, to calculate the square root of a number in cell A1, enter the formula =SQRT(A1) .

What is the square root of 64?

Eight. That's because if we multiply 8 by itself, the result is 64. Taking the square root of a number means finding the number that when multiplied by itself returns the original number.

In the case of the number 64, because its square root is a whole number, 64 is called a perfect square number. This is because it can be represented by arranging 64 items into a square with a side length of 8 items.

a n \huge\sqrt[n]{a} n a ​

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Square Roots

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The square root of a number \(a\) is the answer to the question, \(“\, \)What non-negative number, when squared \((\)raised to the 2\(^\text{nd}\) power \(),\) results in \(a?"\) The symbol for square root is \(“\,\sqrt{\ \ }"\). The square root of the number \(a\) is written as \(\sqrt{a}.\)

What is the square root of \(25?\) Ask yourself the question, \(“\,\)What non-negative number, when squared, results in \(25?"\) The answer to that question will be the square root of \(25\). \(5^2=25\), so \(\sqrt{25}=5\). \(_\square\)

Square roots are crucial in solving quadratic equations and for solving distance problems in geometry.

Definition and Notation

Square roots of perfect squares, estimating square roots of non-perfect squares, solving equations of the form \(x^2=a\), simplifying square roots, rationalizing the denominator, square roots of negative numbers, algorithm for calculating square roots, square roots of complex numbers, problem solving.

The square root of a number \(a\), denoted \(\sqrt{a}\), is the number \(b\) such that \[b^{2} = a\text{ and }b\ge 0.\]

The square root symbol "\(\sqrt{\ }\)" is also sometimes called a radical . The number or expression underneath the top line of the square root symbol is called the radicand. For example, in the expression \(\sqrt{2x+3}\), the radicand is \(2x+3\).

The square root symbol acts as a grouping symbol , which means that all numbers and operations in the radicand are grouped as if they were in parentheses .

Is \(“\,\sqrt{9}+16=\sqrt{9+16}"\) a true statement? ANSWER When calculating \(\sqrt{9}+16\), note that only the \(9\) is in the radicand. Thus, \(\sqrt{9}+16=3+16=19\). When calculating \(\sqrt{9+16}\), note that the sum of \(9\) and \(16\) is in the radicand. Thus, \(\sqrt{9+16}=\sqrt{25}=5\). These values are not equal, and so "\(\sqrt{9}+16=\sqrt{9+16}\)" is a false statement. \(_\square\)

Also, note that the result of a square root operation is always positive or zero . This fact is often ignored and leads to a common misconception about square roots .

The square root of a positive perfect square is always a positive integer. These square roots can be found by thinking of perfect squares until a match for the radicand is found.

For reference, a table of the first ten perfect squares is listed below:

\[\begin{array}{r|r} n & n^2 \\ \hline 0 & 0 \\ 1 & 1 \\ 2 & 4 \\ 3 & 9 \\ 4 & 16 \\ 5 & 25 \\ 6 & 36 \\ 7 & 49 \\ 8 & 64 \\ 9 & 81 \\ 10 & 100 \\ \end{array}\]

Find the value of \(\sqrt{36}\). ANSWER The goal is to think of a non-negative number that, when squared, results in \(36\). Since \(36 = 6^2\), it follows that \(\sqrt{36} = 6.\) \(_\square\)

The square root of a fraction can be found in the same way, as long as both numerator and denominator are perfect squares.

Find the value of \(\sqrt{\dfrac{25}{49}}\). ANSWER The goal is to think of a non-negative number that, when squared, results in \(\dfrac{25}{49}\). Since \(25 = 5^2\) and \(49=7^2\), it follows that \(\sqrt{\dfrac{25}{49}} = \dfrac{5}{7}\). \(_\square\)

The expression

\[ \sqrt{ 2 \frac{1}{4} } + \sqrt{ 2 \frac{7}{9} } \]

can be expressed as \( \frac{a}{b} \), where \(a\) and \(b\) are coprime positive integers. What is \(a+b?\)

If \(x=144\) and \(y=169\), find the value of \(\big(\sqrt{x}+\sqrt{y}\big)^2\).

The square root operation is not just defined for perfect squares. The square root operation can also be applied to any non-negative real number (this domain will later be expanded to negative real numbers and complex numbers ).

When the square root operation is performed on a positive integer that is not a perfect square, the result is an irrational number . Calculators are often used to find the decimal approximation of such a result. However, a decent approximation can be found without a calculator. First, it will be explored how to find integer bounds for a square root.

Which consecutive integers is \(\sqrt{56}\) between? ANSWER First, note that \(56\) is not a perfect square. This means that \(\sqrt{56}\) is an irrational number. What perfect squares is \(56\) between? Looking at the list of perfect squares, \(56\) is between \(49\) and \(64\). Because \(49<56<64\), it stands to reason that \(\sqrt{49}<\sqrt{56}<\sqrt{64}\). Thus, \(7<\sqrt{56}<8\). \(_\square\) This approximation is confirmed when checking on a calculator: \(\sqrt{56}\approx 7.483315\).

This process can be applied to give an integer approximation to any square root. In order to develop a better approximation, one must consider the concavity of the square root function .

To develop a better approximation of a square root, consider which perfect square the number is closer to.

Approximate \(\sqrt{6}\) to one decimal place. ANSWER Note that \(6\) is not a perfect square. Find the integer bounds by considering what perfect squares \(6\) is between. \(6\) is between \(4\) and \(9\). Thus, \(2<\sqrt{6}<3\). Now consider which perfect square 6 is closer to: \(4\) or \(9\). \(6\) is nearly halfway between \(4\) and \(9\), but it is slightly closer to \(4\). Since \(6\) is about halfway between \(4\) and \(9\), it stands to reason that \(\sqrt{6}\) is about halfway between \(2\) and \(3\). \(2.5\) is a good guess to start with. You can check your guess by squaring \(2.5\). The result should be close to \(6\). Multiplying by hand, \(2.5\times 2.5=6.25\). This appears to be close, but is it the best approximation to one decimal place? \(2.5\) appears to be a high guess (since \(6.25>6\), so another valid guess would be \(2.4\). Once again, square this number and compare the result to \(6\). Multiplying by hand, \(2.4\times 2.4=5.76\). Note that this is a low guess, so there are no better one-decimal-place approximations for \(\sqrt{6}\) than \(2.4\) and \(2.5\). Which approximation is better? Note that \(2.4^2=5.76\) is less than \(6\) by \(0.24\), and \(2.5^2=6.25\) is greater than \(6\) by \(0.25\). The closer squared result came from the \(2.4\) approximation. Therefore, \(2.4\) is the best one-decimal-place approximation for \(\sqrt{6}\). This approximation is confirmed with a calculator: \(\sqrt{6}\approx 2.449490\). \(_\square\)

Below is a graph of the square root function, \(f(x)=\sqrt{x}\):

The points with integer coordinates are shown. These points represent \(x\)-values that are perfect squares. Notice that the points become further apart as the function moves further in the positive \(x\) direction.

Taking a closer look, it is more clear what is happening:

In the above graph, the red dashed segments connect the perfect square points. Note that the square root function is always higher than these straight-line segments. What this means is that the value of a square root will tend to be higher than any linear approximation.

Approximate \(\sqrt{2.5}\) to one decimal place. ANSWER \(2.5\) is exactly halfway between the perfect squares \(1\) and \(4\). Using linear thinking, it would stand to reason that the best approximation for \(\sqrt{2.5}\) would be exactly halfway between \(1\) and \(2\). The guess becomes \(1.5\). Multiplying by hand, \(1.5\times 1.5=2.25\). The guess is low, it is \(0.25\) off from the value of \(2.5\). Recall that the actual square root tends to be greater than a linear approximation. Now try a guess of \(1.6\). Multiplying by hand, \(1.6\times 1.6=2.56\). This guess is high, and it is only off by \(0.06\). Thus, the best one-decimal-place approximation for \(\sqrt{2.5}\) is \(1.6\). It turns out that the best approximation is not halfway between \(1\) and \(2\). This approximation is confirmed by calculator: \(\sqrt{2.5}\approx 1.581139\). \(_\square\)

Let \(a\) and \(b\) be distinct non-negative integers, and let \(x=\frac{a^2+b^2}{2}\).

Which of these statements is true about \(\sqrt{x}\)?

Square roots are often used to find the solutions to quadratic equations . The simplest kind of quadratic equation that can be solved with square roots is an equation of the form \(x^2=a\).

Find the solutions to \(x^2=16.\) ANSWER The square root can be used to find a solution to this equation. This solution is \(x=\sqrt{16}=4\). However, there is another solution. This other solution is \(x=-\sqrt{16}=-4\). \(_\square\)

An equation of the form \(x^2=a\), where \(a>0\), will always have two solutions: one positive and one negative.

Let \(a\) be a positive real number . The solutions to the equation \(x^2=a\) are \[x=\sqrt{a}\quad \text{and} \quad x=-\sqrt{a},\] or equivalently, \[x=\pm \sqrt{a}.\]

For these kinds of equations, the positive solution is called the principle square root , while the negative solution is called the negative square root .

Some problems will request, "What are the square roots of a certain number, \(a\)?" Normally, we would consider \(\sqrt{a}\) to have only one possible result. However, in this context, the meaning of the question is to find both solutions of the equation \(x^2=a\).

The square root of a positive integer that is not a perfect square is always an irrational number . The decimal representation of such a number loses precision when it is rounded, and it is time-consuming to compute without the aid of a calculator . Instead of using decimal representation, the standard way to write such a number is to use simplified radical form :

Let \(a\) be a positive non-perfect square integer. The simplified radical form of the square root of \(a\) is \[b\sqrt{c}.\] In this form \(\sqrt{a}=b\sqrt{c}\), both \(b\) and \(c\) are positive integers, and \(c\) contains no perfect square factors other than \(1\).

The process for putting a square root into simplified radical form involves finding perfect square factors, then applying the following identity:

Let \(a\) and \(b\) be positive real numbers. Then, \[\sqrt{ab}=\sqrt{a}\times\sqrt{b}.\]

Simplify \(\sqrt{72}\). ANSWER First, ask yourself, "What is a perfect square factor of \(72\)?" \(4\) is a perfect square factor of \(72\), and \(9\) is a perfect square factor of \(72\). For the sake of this process, it is more efficient to find the largest perfect square factor of \(72\). As shown below, \(4\times 9=36\) is the largest perfect square factor of \(72\): \[\begin{align} \sqrt{72}&=\sqrt{36\times 2} \\ &=\sqrt{36}\times\sqrt{2} \\ &=6\sqrt{2}. \end{align}\] Therefore, simplified form of \(\sqrt{72}\) is \(6\sqrt{2}.\ _\square\) Note : When a number is placed to the left of a square root symbol, multiplication is implied. "\(6\sqrt{2}\)" is read as "\(6\) times the square root of \(2\)."

Rationalizing the denominator is the process of re-writing a rational expression that contains a radical in the denominator as an equivalent rational expression that contains no radical in the denominator.

The following identity is a direct consequence of the definition of a square root, and it is crucial in the process of rationalizing a denominator:

Let \(a\) be a positive real number. Then, \[\sqrt{a}\times\sqrt{a}=a.\]

Rationalizing the denominator involves applying the identity property of multiplication: the fact that multiplying the numerator and denominator by the same number results in an equivalent rational expression. In this case, the number we choose for this is the square root.

Let \(a\) be a positive real number. In order to rationalize the fraction \(\dfrac{1}{\sqrt{a}}\), both the numerator and the denominator must be multiplied by \(\sqrt{a}\). This is equivalent to multiplying the expression by \(1\), so the resulting expression is equivalent to \(\dfrac{1}{\sqrt{a}}\): \[\begin{align} \frac{1}{\sqrt{a}}&= \frac{1}{\sqrt{a}} \times \frac{\sqrt{a}}{\sqrt{a}} \\ &= \frac{1 \times \sqrt{a}}{\sqrt{a} \times \sqrt{a}} \\ &= \frac{\sqrt{a}}{a}. \end{align}\]

Rationalize the following expression:\[\] \[\frac{1}{\sqrt{2}}.\] ANSWER Rationalization will help in calculation as the denominator will be an integer: \[\begin{align} x &=\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \\ &=\frac{\sqrt{2}}{2}. \ _\square \end{align}\]

Rationalize the denominator of \(\dfrac{1}{\sqrt{\sqrt{5}}}\). ANSWER We have \[\begin{align} \dfrac{1}{\sqrt{\sqrt{5}}} & = \dfrac{1 × \sqrt{\sqrt{5}}}{\sqrt{\sqrt{5}} × \sqrt{\sqrt{5}}}\\ \\ & = \dfrac{\sqrt{\sqrt{5}}}{\sqrt{5}}\\ \\ & = \dfrac{\sqrt{\sqrt{5}} × \sqrt{5}}{\sqrt{5} × \sqrt{5}}\\ \\ & = \dfrac{\sqrt{5\sqrt{5}}}{5}.\ _\square \end{align}\]

Sometimes, you will encounter denominators in which an unfavorable term is added to another, making the two seemingly inseparable. In these scenarios, you must multiply the conjugate of the number.

The conjugate of a binomial number \((a+b)\) is the number \((a-b)\). When multiplied to the original number, it will produce \(\displaystyle a^2 - b^2\).

It can be proven rather simply that this always works:

We have \[\begin{align} (a+b) \times (a-b) &=a^2 - ab + ab - b^2 \\ &=a^2 - b^2. \end{align}\]

We can use this technique of multiplying by the conjugate to rationalize denominators with addition or subtraction in them.

Rationalize the following expression: \[\dfrac{1}{\sqrt{3} - 1}.\] ANSWER To rationalize this, multiply by the conjugate: \[\begin{align} x &=\frac{1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} \\ &=\frac{\sqrt{3} + 1}{2}. \ _\square \end{align}\]

Rationalize the following expression and find the value of a + b: \[\dfrac{\sqrt{3} - 1}{\sqrt{3} + 1} = {a + b \sqrt{3}}.\] ANSWER To rationalize this, multiply by the conjugate: \[\begin{align} x &=\frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} \\ &=\frac{(\sqrt{3} - 1)^2}{2}\\ &=\frac{4-2\sqrt{3}}{2}\\ &= 2-\sqrt{3} \\ &= a + b \sqrt{3}\\\\ \Rightarrow a+b& =2-1 \\ &= 1. \ _\square \end{align}\]

The rationalization of the expression

\[\dfrac{1}{\sqrt{5} + \sqrt{2} + 1}\]

can be expressed as

\[\dfrac{- A\sqrt{10} - B\sqrt{5} + C\sqrt{2} + D}{E},\]

where \(A\), \(B\), \(C\), \(D\), and \(E\) are all positive integers.

Find the minimum value of \(A + B + C + D + E\).

This problem is part of the set "Xenophobia."

You may have noticed that none of these problems or examples involve finding the square root of a negative number.

Find the value of \(\sqrt{-9}\). ANSWER What is a non-negative number that, when squared, results in \(-9?\) It is not \(3\), because \(3^2\) results in positive \(9\). Even if we allowed ourselves to answer with a negative number, we still would not find a good result. It cannot be \(-3\), because \((-3)^2\) results in positive \(9\). It turns out that there is no real number value for \(\sqrt{-9}\). Evaluating \(\sqrt{-9}\) requires the imaginary unit , \(i\). Because \(i^2=-1\), \((3i)^2=-9\). Thus, \(\sqrt{-9}=3i\). \(_\square\)

The imaginary unit , \(i\), is a complex number that satisfies the following equation: \[i^2=-1.\]

With the introduction of complex numbers, the square root of any negative number can be evaluated:

Let \(a\) be a positive real number. Then, \[\sqrt{-a}=i\sqrt{a}.\]

\[\large \sqrt{-2}\times\sqrt{-3}=\, ?\]

In this problem, the square root is a function from the complex numbers to the complex numbers.

I am performing the following steps to prove \(-1 = 1:\)

• \(-1 = i^{2}\)
• \(-1=i \times i\)
• \(-1 = \sqrt{-1} \times \sqrt{-1}\)
• \(-1 =\sqrt{-1 \times -1}\)
• \(-1 =\sqrt{1}\)

In which step does the error first appear?

The square root of a non-negative, non-perfect square real number is always an irrational number . Finding the decimal representation of such a square root will often be accomplished with the help of a calculator. However, you may sometimes be asked to manually calculate the square root of some non-perfect square number.

Find the square root of 39 to 3 decimal places. ANSWER We have \[ \begin{array} {c|ccccc|c} 6 & \overline{39}. & \overline{00} & \overline{00} & \overline{00} & \overline{00} & \underline{6.2449\ldots} \\ & \underline{- 36} & & & & & \\ 122 & 3 & 00 & & & & \\ & \underline{- 2} & \underline{44} & & & & \\ 1244 & & 56 & 00 & & & \\ & & \underline{- 49} & \underline{76} & & & \\ 12484 & & 6 & 24 & 00 & & \\ & & \underline{4} & \underline{99} & \underline{36} & & \\ 124889 & & 1 & 24 & 64 & 00 & \\ & & \underline{1} & \underline{12} & \underline{40} & \underline{01} & \\ & & & 12 & 23 & 99 & \\ \end{array}\] Thus, the square root of 39 to 3 decimal places is 6.245. \(_\square\)

Checking on a calculator, you find the answer to be correct. But how is this so? What was the method? All it looked like was a bunch of random numbers. Well, the method resembles long division, except you bring two digits down instead of one after subtracting. Here is the method:

If there is 1 digit, the digit will be by itself. If there are 2 digits, the digits can be paired together.

Now, from the most leftward pair of digit, or pair of digits, find the largest positive integer that has its square less than this 1 or 2-digit number. In the example above, the largest integer satisfying this was 6, since \(6^2 < 39 < 7^2\). So, we put the 6 in the divisor column, and put a 6 in the quotient column.

We subtract the square of this number from the first pair of digits, then drop the next pair down. In the example above, we subtracted \(39-36\) to get \(3\), then drop \(00\) from the top of the column to get \(300\).

We add the number from the divisor to itself, then shift it across to the left. We now have to find an integer \(a\) such that \(\overline{ka} \times a < \text{New dividend} < \overline{k(a+1)} \times {a+1}\), where \(k\) refers to the number obtained after adding the old divisor to itself and \(\overline{abc}\) refers to concatenating the numbers \(a\), \(b\) and \(c\). Once we find the value for \(a\), we write it down in place, put it as the next number in the quotient. E.g., above, we had \(\overline{12a}\times a = 300\), so \(a=2\) since \(122 \times 2 < 300 < 123 \times 3\). Now, we have got the first digit of the square root. Now, we add the first divisor and the first digit. Then we have to find a number \(a\) such that \(\overline{ka} × a \leq\) the present remainder where \(k\) is the second dividend. For example in the example above, the first step is as follows: \[\begin{array} {c|ccccc|c} \color{blue}{6} & \overline{39}. & \overline{00} & \overline{00} & \overline{00} & \overline{00} & \underline{\color{blue}{6}} \\ & \underline{- 36} & & & & &\\ & 3 & 00 & & & & \end{array}\] Now, we add the blue coloured 6's to get \(\color{purple}{12}\). Then we have to find a number which fits in the box such that \(12\boxed{\phantom{0}} × \boxed{\phantom{0}} \leq 300\). We see that \(12\color{green}{2} × \color{green}{2} = 244 < 300\) and \(12\color{orange}{3} × \color{orange}{3} = 369 > 300\). So, we take \(2\) in the box, i.e. \(a = 2\). So, we finally have \[ \begin{array} {c|ccccc|c} 6 & \overline{39}. & \overline{00} & \overline{00} & \overline{00} & \overline{00} & \underline{6.\color{orange}{2}} \\ & \underline{- 36} & & & & & \\ \color{purple}{12}\color{red}{2} & 3 & 00 & & & & \\ & \underline{- 2} & \underline{44} & & & & \\ & & 56 & & & & & \end{array}\]

We find the \(\overline{ka} \times a\), and then subtract this from the new dividend, then drop the next pair down. E.g., above, we had \(300-244=56\), so the next new dividend became \(5600\).

Add the number \(a\) to \(\overline{ka}\), and shift to the left again. Repeat Steps 4-6 until all the pairs have been exhausted. You have now determined the square root manually.

Though the method given above is good, it is also slightly lengthy if many decimal places are required. When the number of significant figures takes priority, we need a better algorithm.

Proposed is the Babylonian method to compute \(x=\sqrt{\alpha}\) :

Pick a number \(x_0\) such that \(x_0 > \sqrt{\alpha}\).

Define a sequence \(\left \{x_n \right \}_{n=0}^{\infty}\) that is built recursively by the following formula: \[x_{n+1} = \frac{1}{2} \left( x_n + \frac{\alpha}{x_n} \right).\]

\(\displaystyle \sqrt{\alpha} = \lim_{n \to \infty} x_n\).

Implementation in Python :

Code in action :

Compare with the actual value \(\sqrt{39} = 6.2449979983983982058468...\); this algorithm returns a number that is correct up to 16 significant figures, with only 7 iterations.

We can get the approximate value of \(\sqrt{\alpha}\) to eight to sixteen significant figures in 3 to 4 iterations if we chose \(x_0\) suitably.

This is a result of the following inequality, whose proof is given below:

If \(\varepsilon_n = x_n - \sqrt{\alpha}\) and \(\beta=2\sqrt{\alpha}\), then

\[0< \frac{\varepsilon_n}{\beta} < \left(\frac{\varepsilon_0}{\beta} \right)^{2^n} \quad n = 1,2, \ldots. \]

So, if \(\varepsilon_0 < \beta\) \(\Big(\)prefrerably \(\varepsilon_0 < \frac{\beta}{10}\Big) \), then the error decreases rapidly. Usually, to do this, taking \(x_0=\left\lceil \sqrt{\alpha} \right\rceil\) is sufficient. Here \(\lceil \cdot \rceil\) denotes the ceiling function .

We have \[ x_{n +1} = \frac{1}{2} \left(x_n + \frac{\alpha}{x_n}\right). \] We first show \(x_{n} > \sqrt{\alpha}\) for any \(n \in \mathbb{N} \). By the AM-GM inequality, \[ x_{n} > \frac{1}{2} \left(2 \sqrt{x_{n - 1} \cdot \frac{\alpha}{x_{n - 1}}}\right) = \sqrt{\alpha}. \] We cannot have equality as \(x_0 > \sqrt{\alpha} \). From here, it is easy to see that if \(x_0 = \sqrt{\alpha}, \) then \(x_n\) are all equal for all integers \(n\). Now, \[\begin{align} x_{n + 1} - \sqrt{\alpha} &= \frac{1}{2} \left(x_n + \frac{\alpha}{x_n} - 2\sqrt{\alpha} \right) \\ &= \frac{1}{2} \left(\sqrt{x_n} - \sqrt{\frac{\alpha}{x_n}} \right)^{2} \\ &= \frac{\left(x_n - \sqrt{\alpha} \right)^{2}}{2 x_n}. \end{align}\] So, \[ \frac{\varepsilon_{n + 1}}{\beta} = \frac{\left(x_n - \sqrt{\alpha} \right)^{2}}{4 x_n \sqrt{\alpha}} < \frac{\left(x_n - \sqrt{\alpha} \right)^{2}}{4 \alpha} = \left(\frac{\varepsilon_{n}}{\beta}\right)^2,\] and by induction, it is easy to see that \[\frac{\varepsilon_{n}}{\beta}< \left(\frac{\varepsilon_{0}}{\beta}\right)^{2^n}\] for \( n > 0 \). Note that \(\varepsilon_n > 0 \) as \( x_{n} > \sqrt{\alpha} \), then \(\dfrac{\varepsilon_n}{\beta} > 0.\) This completes the proof. \(_\square\)

This method is illustrated by the following examples:

Compute \(\sqrt{39}\) correct to six decimal places. ANSWER Let us take \(x_0 = 7\). Let's figure out the value of \(\dfrac{\varepsilon_0}{\beta}\): \[\begin{align} \frac{\varepsilon_0}{\beta} &= \frac{7-\sqrt{39}}{2\sqrt{39}} \\ &< \frac{1}{2\sqrt{39}} \\ &< \frac{1}{12} \\\implies 0 &< \frac{\varepsilon_0}{\beta} < \frac{1}{10}. \end{align} \] Therefore, we need to calculate just three iterates. This is because of the following inequality \((\)note that \(\beta<14)\): \[\boxed{ 0 < \varepsilon_3 < 14 \times 10^{-8} < 10^{-6} }. \] Iterating, we get \[\begin{align} x_1 &= 6.28571428\ldots \\ x_2 &= 6.24512987\ldots \\ x_3 &= 6.24499799\ldots. \end{align}\] Hence the value of \(\sqrt{39}\), correct to six decimals, is 6.244998. \(_\square\) Note: Notice that right after two iterates, we get the value of \(\sqrt{39}\) correct to three decimals, compared to three iterates required in the previous method. This method requires only three iterates, compared to 6 iterates required by the previous method for this task.

Compute \(\sqrt{3}\) correct to thirty decimal places. ANSWER Let us take \(x_0 = 2\). Let's calculate \(\dfrac{\varepsilon_0}{\beta}\): \[\begin{align} \frac{\varepsilon_0}{\beta} &= \frac{2-\sqrt3}{2\sqrt3} \\ &= \frac{1}{2\sqrt3(2+\sqrt3)} \\ &= \frac{1}{2(2\sqrt3+3)} \\ &< \frac{1}{2(5)} \\\implies 0&<\frac{\varepsilon_0}{\beta} < \frac{1}{10}. \end{align} \] This means that \[\boxed{0< \varepsilon_5 < 4 \times 10^{-32}}\] since \(\beta < 4.\) Hence we need to calculate just 5 iterates to get \(\sqrt3\) correct to thirty decimals ! Iterating, we get (only ten digits are shown here) \[\sqrt3 \approx x_5 = 1.7320508075\ldots.\ _\square\] Note: With 20 iterates, the value of \(\sqrt3\) to more than a million digits is known!

For more details in approximating square roots, see here .

The square root of a complex number is somewhat ambiguous. Non-real complex numbers are neither positive nor negative, so it is not well-defined which square root is the principal square root. Therefore, when the square root operation is used on a complex number, the result is interpreted to be all the solutions of an equation:

Let \(z\) be a complex number. Then there are up to two values for \(\sqrt{z}\), and they are equal to the solutions of the equation \[x^2=z.\]

Note that the square root operation, when used on complex numbers, is not well-defined in the sense that there is only one result.

Find square root of \(i\) using general method. Assume that \(\sqrt{i}=a+ib\). Then, \(i=(a+ib)^{2}=a^{2}-b^{2}+2abi\). Equating the real parts of the equation and equating the imaginary parts of the equation gives \[\begin{array}a^{2}-b^{2}=0 & \text{and} & 2ab=1\end{array}.\] Solving this system of equations gives \[\begin{array}a=b=\frac{1}{\sqrt{2}} & \text{or} & a=b=-\frac {1}{\sqrt{2}}.\end{array}\] Thus, \[\begin{array}\sqrt{i}=\dfrac {1}{\sqrt{2}}+ \dfrac {i}{\sqrt{2}} & \text{or} & \sqrt{i}=-\dfrac {1}{\sqrt{2}}- \dfrac {i}{\sqrt{2}}.\ _\square \end{array}\]

In general, for complex number \(z=a+ib\) such that \(a,b\in\mathbb{R}\), \[\sqrt{z}=\pm \left[ \sqrt {\frac{\sqrt{a^2+b^2}+a}{2} }+i \sqrt{ \frac{\sqrt{a^2+b^2}-a}{2} } \right]. \]

Another method, using Euler's formula , is to express the complex number \(z\) in the form \(z=re^{i\theta},\) where \(r=|z|\) and \(\theta \in [0, 2\pi) \). Here's an illustration:

Find the square root of \(i\) using Euler's formula. Let \(z=i\) so that \(r=|z|=|i|=1\). Since \(e^{i\pi/2}=\cos \frac{\pi}{2} + i\sin \frac{\pi}{2} = i \), we have \(\theta= \frac{\pi}{2}. \) Now, \[\sqrt{z}=\pm \sqrt{r} e^{i\theta/2}. \] Hence, with \(z=i\), we get \(\displaystyle\sqrt{i}=\pm e^{i\pi/4}=\pm \left(\frac {1}{\sqrt{2}}+ \frac {i}{\sqrt{2}}\right),\) which is the same answer as before, but much faster to calculate. \(_\square\)

Any complex number \(z=a+bi\) can be written in the polar form as

\[z = r(\cos \theta + i \sin \theta),\]

where \(r=\sqrt{a^2+b^2}\) and \(\theta=\arctan{\frac{b}{a}}\). Consider the square root by raising its power to half and apply De Moivre's theorem :

\[\begin{aligned} z^\frac{1}{2} &= \pm r^\frac{1}{2}(\cos \theta + i \sin \theta)^\frac{1}{2} \\ \sqrt{z} &= \pm \sqrt{r}\left(\cos{\frac{\theta}{2}} + i\sin{{\frac{\theta}{2}}}\right). \end{aligned}\]

\[(a + ib)^2 = 7 + 24i \]

If \(a\) and \(b\) are integers satisfying the equation above, then find the value of \( |a| - |b| \).

• \(i = \sqrt{-1}\) is the imaginary unit.
• \(| \cdot |\) is the absolute value function.

\[A=\sqrt{65}-8, \quad B=\sqrt{50}-7 \]

Azhaghu was once learning square roots as he liked to compare radical expressions. He worked day and night on his new topic and found the two expressions above.

After working hard, reading Naruto and eating chips, Azhaghu figured out correctly which of \(A\) and \(B\) is bigger.

So which is indeed bigger?

Bonus : Generalize this.

Find \(a\), where\[\] \[a = \dfrac{1}{\sqrt{1} + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + \cdots + \dfrac{1}{\sqrt{99} + \sqrt{100}}.\] ANSWER Similar to the previous problem, multiply by the conjugate: \[\begin{align} a =&\dfrac{1}{\sqrt{1} + \sqrt{2}} \times \frac{-\sqrt{1}+\sqrt{2}}{-\sqrt{1}+\sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} \times \frac{-\sqrt{2}+\sqrt{3}}{-\sqrt{2}+\sqrt{3}} \\\\&+ \dfrac{1}{\sqrt{3} + \sqrt{4}} \times \frac{-\sqrt{3}+\sqrt{4}}{-\sqrt{3}+\sqrt{4}} + \cdots + \dfrac{1}{\sqrt{99} + \sqrt{100}} \times \frac{-\sqrt{99}+\sqrt{100}}{-\sqrt{99}+\sqrt{100}}\\\\ =&\frac{-\sqrt{1}+\sqrt{2}}{1} + \frac{-\sqrt{2}+\sqrt{3}}{1} + \frac{-\sqrt{3}+\sqrt{4}}{1} + \cdots + \frac{-\sqrt{99}+\sqrt{100}}{1}\\\\ =& - \sqrt{1} + \sqrt{2} - \sqrt{2} + \sqrt{3} - \sqrt{3} + \sqrt{4} - \cdots - \sqrt{99} + \sqrt{100}\\\\ =& -\sqrt{1} + \sqrt{100}\\\\ =&-1 + 10\\\\ =&9.\ _\square \end{align}\]

True or False?

If \(x\) and \(y\) are real numbers such that \( \sqrt{x} = \sqrt{y} \), then

\[ x = y. \]

Given that 1234321 is a perfect square, quickly find the value of

\[\sqrt{1234321}. \]

What is the value of

\[ \large \sqrt{1\text{%}}\, ? \]

Using the fact that \((\sqrt{x}+\sqrt{y})^2=x+y+2\sqrt{xy}\), find the square root of \(5+\sqrt{24}\). This number can be expressed in the form \(\sqrt{a}+\sqrt{b}\), where \(a\leq{b}\).

Find the value of \(b-a\).

\[\large \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2 + \cdots}}}} = \, ?\]

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Solving Quadratics by Square Root Method

How to solve quadratic equations using the square root method.

This is the “best” method whenever the quadratic equation  only contains [latex]{x^2}[/latex] terms. That implies no presence of any [latex]x[/latex] term being raised to the first power somewhere in the equation.

The general approach is to collect all [latex]{x^2}[/latex] terms on one side of the equation while keeping the constants to the opposite side. After doing so, the next obvious step is to take the square roots of both sides to solve for the value of [latex]x[/latex]. Always attach the [latex] \pm [/latex] symbol when you get the square root of the constant.

Examples of How to Solve Quadratic Equations by Square Root Method

Example 1 : Solve the quadratic equation below using the Square Root Method.

I will isolate the only [latex]{x^2}[/latex] term on the left side by adding both sides by [latex] + 1[/latex]. Then solve the values of [latex]x[/latex] by taking the square roots of both sides of the equation. As I mentioned before, we need to attach the plus or minus symbol to the square root of the constant.

So I have [latex]x = 5[/latex] and [latex]x = – \,5[/latex] as final answers since both of these values satisfy the original quadratic equation. I will leave it to you to verify.

Example 2 : Solve the quadratic equation below using the Square Root Method.

This problem is very similar to the previous example. The only difference is that after I have separated the [latex]{x^2}[/latex] term and the constant in the opposite sides of the equation, I need to divide the equation by the coefficient of the squared term before taking the square roots of both sides.

The final answers are [latex]x = 4[/latex] and [latex]x = – \,4[/latex].

Example 3 : Solve the quadratic equation below using the Square Root Method.

I can see that I have two [latex]{x^2}[/latex] terms, one on each side of the equation. My approach is to collect all the squared terms of [latex]x[/latex] to the left side, and combine all the constants to the right side. Then solve for [latex]x[/latex] as usual, just like in Examples 1 and 2.

The solutions to this quadratic formula are [latex]x = 3[/latex] and [latex]x = – \,3[/latex].

Example 4 : Solve the quadratic equation below using the Square Root Method.

The two parentheses should not bother you at all. The fact remains that all variables come in the squared form, which is what we want. This problem is perfectly solvable using the square root method.

So my first step is to eliminate both of the parentheses by applying the distributive property of multiplication. Once they are gone, I can easily combine like terms. Keep the [latex]{x^2}[/latex] terms to the left, and constants to the right. Finally, apply square root operation in both sides and we’re done!

Not too bad, right?

Example 5 : Solve the quadratic equation below using the Square Root Method.

Since the [latex]x[/latex]-term is being raised to the second power twice, that means, I need to perform two square root operations in order to solve for [latex]x[/latex].

The first step is to have something like this: (   ) 2 = constant . This allows me to get rid of the exponent of the parenthesis on the first application of square root operation.

After doing so, what remains is the “stuff” inside the parenthesis which has an [latex]{x^2}[/latex] term. Well, this is great since I already know how to handle it just like the previous examples.

There’s an [latex]x[/latex]-squared term left after the first application of square root.

Now we have to break up [latex]{x^2} = \pm \,6 + 10[/latex] into two cases because of the “plus” or “minus” in [latex]6[/latex].

• Solve the first case where [latex]6[/latex] is positive .

• Solve the second case where [latex]6[/latex] is negative .

The solutions to this quadratic equations are [latex]x = 4[/latex], [latex]x = – \,4[/latex], [latex]x = 2[/latex], and [latex]x = – \,2[/latex]. Yes, we have four values of [latex]x[/latex] that can satisfy the original quadratic equation.

Example 6 : Solve the quadratic equation below using the Square Root Method.

Example 7 : Solve the quadratic equation below using the Square Root Method.

You might also like these tutorials:

• Solving Quadratic Equations by Factoring Method
• Solving Quadratic Equations by the Quadratic Formula
• Solving Quadratic Equations by Completing the Square

• RCA 101 – 5-Why Analysis (Free Training)
• RCA201 – Basic Failure Analysis
• RCA 301 – PROACT® RCA Certification
• RCA401 – RCA Train The Trainer
• Other Trainings
• 5 Whys Root Cause Analysis Template
• RCA Template
• Chronic Failure Calculator

7 Powerful Root Cause Analysis Tools and Techniques

By Sebastian Traeger

Updated: April 21, 2024

Reading Time: 5 minutes

1. The Ishikawa Fishbone Diagram (IFD)

2. pareto chart, 4. failure mode and effects analysis (fmea), 5. proact® rca method, 6. affinity diagram, 7. fault tree analysis (fta).

With over two decades in business – spanning strategy consulting, tech startups and executive leadership – I am committed to helping your organization thrive. At Reliability, we’re on a mission to help enhance strategic decision-making and operational excellence through the power of Root Cause Analysis, and I hope this article will be helpful!  Our goal is to help you better understand these root cause analysis techniques by offering insights and practical tips based on years of experience. Whether you’re new to doing RCAs or a seasoned pro, we trust this will be useful in your journey towards working hard and working smart.

Root Cause Analysis (RCA) shines as a pivotal process that helps organizations identify the underlying reasons for problems, failures, and inefficiencies. The goal is simple: find the cause, fix it, and prevent it from happening again. But the process can be complex, and that’s where various RCA techniques come into play. 

Let’s dive into seven widely utilized RCA techniques and explore how they can empower your team’s problem-solving efforts.

Named after Japanese quality control statistician Kaoru Ishikawa, the Fishbone Diagram is a visual tool designed for group discussions. It helps teams track back to the potential root causes of a problem by sorting and relating them in a structured way. The diagram resembles a fishbone, with the problem at the head and the causes branching off the spine like bones. This visualization aids in categorizing potential causes and studying their complex interrelationships.

The Pareto Chart, rooted in the Pareto Principle, is a visual tool that helps teams identify the most significant factors in a set of data. In most situations, 80% of problems can be traced back to about 20% of causes. By arranging bar heights from tallest to shortest, teams can prioritize the most significant factors and focus their improvement efforts where they can have the most impact.

The 5 Whys method is the epitome of simplicity in getting to the bottom of a problem. By repeatedly asking ‘why’ (typically five times), you can delve beneath the surface-level symptoms of a problem to unearth the root cause. This iterative interrogation is most effective when answers are grounded in factual evidence.

When prevention is better than cure, Failure Mode and Effects Analysis (FMEA) steps in. This systematic, proactive method helps teams identify where and how a process might fail. By predicting and examining potential process breakdowns and their impacts, teams can rectify issues before they turn into failures. FMEA is a three-step process that involves identifying potential failures, analyzing their effects, and prioritizing them based on severity, occurrence, and detection ratings.

The PROACT ® RCA technique is a robust process designed to drive significant business results. Notably used to identify and analyze ‘chronic failures,’ which can otherwise be overlooked, this method is defined by its name:

PReserving Evidence and Acquiring Data: Initial evidence collection step based on the 5-P’s – Parts, Position, People, Paper, and Paradigms.

Order Your Analysis Team and Assign Resources: Assembling an unbiased team to analyze a specific failure.

Analyze the Event: Reconstructing the event using a logic tree to identify Physical, Human, and Latent Root Causes.

Communicate Findings and Recommendations: Developing and implementing solutions to prevent root cause recurrence.

Track and Measure Impact for Bottom Line Results: Tracking the success of implemented recommendations and correlating the RCA’s effectiveness with ROI.

PROACT® RCA excels in mitigating risk, optimizing cost, and boosting performance, making it a valuable addition to any RCA toolkit.

The Affinity Diagram is a powerful tool for dealing with large amounts of data. It organizes a broad range of information into groups based on their natural relationships, creating a clear, visual representation of complex situations. It’s particularly beneficial for condensing feedback from brainstorming sessions into manageable categories, fostering a better understanding of the broader picture.

Fault Tree Analysis (FTA) is a top-down, deductive failure analysis that explores the causes of faults or problems. It involves graphically mapping multiple causal chains to track back to possible root causes, using a tree-like diagram. FTA is particularly useful in high-risk industries, such as aerospace and nuclear power, where preventing failure is crucial.

Each RCA technique provides a unique approach for viewing and understanding problems, helping you pinpoint the root cause more effectively. The key is to understand when and how to use each tool, which can significantly enhance your team’s problem-solving capabilities.

Power up your RCA analysis with our EasyRCA and revolutionize your problem-solving process. Start Your Free Trial.

In conclusion, the techniques presented offer a diverse set of tools to help organizations address problems and inefficiencies effectively. From visual representations like the Ishikawa Fishbone Diagram and Pareto Chart to more proactive approaches such as the 5 Whys and Failure Mode and Effects Analysis (FMEA), each technique provides a unique perspective on identifying and mitigating root causes.

The PROACT® RCA Method stands out for its comprehensive process, particularly suited for chronic failures. Additionally, the Affinity Diagram and Fault Tree Analysis (FTA) contribute valuable insights by organizing data and exploring causal chains, respectively. Leveraging these techniques strategically enhances a team’s problem-solving capabilities, enabling them to make informed decisions and drive continuous improvement.

I hope you found these 7 techniques insightful and actionable! Stay tuned for more thought-provoking articles as we continue to share our knowledge. Success is rooted in a thorough understanding and consistent application, and we hope this article was a step in unlocking the full potential of Root Cause Analysis for your organization. Reliability runs initiatives such as an online learning center focused on the proprietary PROACT® RCA methodology and EasyRCA.com software. For additional resources, visit our Reliability Resources .

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9.1: Extracting Square Roots

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Learning Objectives

• Solve quadratic equations by extracting square roots.

Extracting Square Roots

Recall that a quadratic equation is in standard form if it is equal to 0:

\[a x^{2}+b x+c=0\]

where a , b , and c are real numbers and \(a\neq 0\). A solution to such an equation is called a root . Quadratic equations can have two real solutions, one real solution, or no real solution. If the quadratic expression on the left factors, then we can solve it by factoring. A review of the steps used to solve by factoring follow:

Step 1: Express the quadratic equation in standard form.

Step 2: Factor the quadratic expression.

Step 3: Apply the zero-product property and set each variable factor equal to 0.

Step 4: Solve the resulting linear equations.

For example, we can solve \(x^{2}-4=0\) by factoring as follows:

The two solutions are −2 and 2. The goal in this section is to develop an alternative method that can be used to easily solve equations where b = 0, giving the form

\[a x^{2}+c=0\]

The equation \(x^{2}-4=0\) is in this form and can be solved by first isolating \(x^{2}\).

If we take the square root of both sides of this equation, we obtain the following:

\(\begin{aligned} \sqrt{x^{2}} &=\sqrt{4} \\|x| &=2 \end{aligned}\)

Here we see that \(x=-2\) and \(x=2\) are solutions to the resulting equation. In general, this describes the square root property ; for any real number k ,

\[\text{If}\:x^{2}=k, \text { then } x=\pm \sqrt{k}\]

The notation “±” is read “plus or minus” and is used as compact notation that indicates two solutions. Hence the statement \(x=\pm \sqrt{k}\) indicates that \(x= - \sqrt{k}\)or \(x= + \sqrt{k}\). Applying the square root property as a means of solving a quadratic equation is called extracting the roots .

Example \(\PageIndex{1}\)

\(x^{2}-25=0\)

Begin by isolating the square.

Next, apply the square root property.

\(\begin{aligned} x^{2} &=25 \\ x &=\pm \sqrt{25} \\ x &=\pm 5 \end{aligned}\)

The solutions are −5 and 5. The check is left to the reader.

Certainly, the previous example could have been solved just as easily by factoring. However, it demonstrates a technique that can be used to solve equations in this form that do not factor.

Example \(\PageIndex{2}\)

\(x^{2}-5=0\)

Notice that the quadratic expression on the left does not factor. We can extract the roots if we first isolate the leading term, \(x^{2}\).

Apply the square root property.

\(x=\pm \sqrt{5}\)

For completeness, check that these two real solutions solve the original quadratic equation. Generally, the check is optional.

\(\begin{array}{rlrl}{\text { Check } x=-\sqrt{5}} & {\text { Check } x=\sqrt{5}} \\ {x^{2}-5=0} & {x^{2}-5=0} \\ {(\color{OliveGreen}{-\sqrt{5}}\color{black}{)}^{2}-5=0} & {(\color{OliveGreen}{\sqrt{5}}\color{black}{)}^{2}-5=0} \\ {5-5=0} & {5-5=0} \\ {0=0 \color{Cerulean}{\checkmark}} & {0=0\color{Cerulean}{\checkmark}}\end{array}\)

Answer : The solutions are \(-\sqrt{5}\) and \(\sqrt{5}\).

Example \(\PageIndex{3}\)

Begin by isolating \(x^{2}\).

Apply the square root property and then simplify.

\(\begin{array}{l}{x=\pm \sqrt{\frac{45}{4}}} \\ {x=\pm \frac{\sqrt{9 \cdot 5}}{\sqrt{4}}} \\ {x=\pm \frac{3 \sqrt{5}}{2}}\end{array}\)

The solutions are \(-\frac{3 \sqrt{5}}{2}\) and \(\frac{3 \sqrt{5}}{2}\).

Sometimes quadratic equations have no real solution.

Example \(\PageIndex{4}\)

\(x^{2} + 9 =0\)

After applying the square root property, we are left with the square root of a negative number. Therefore, there is no real solution to this equation.

No real solution

Reverse this process to find equations with given solutions of the form ±k .

Example \(\PageIndex{5}\)

Find an equation with solutions \(-2 \sqrt{3}\) and \(2 \sqrt{3}\).

Begin by squaring both sides of the following equation:

\(\begin{array}{l}{x=\pm 2 \sqrt{3}} \\ {x^{2}=( \pm 2 \sqrt{3})^{2}} \\ {x^{2}=4 \cdot 3} \\ {x^{2}=12}\end{array}\)

Lastly, subtract 12 from both sides and present the equation in standard form.

\(x^{2}-12=0\)

Exercise \(\PageIndex{1}\)

\(9x^{2}-8=0\).

\(x=-\frac{2 \sqrt{2}}{3}\) or \(x=\frac{2 \sqrt{2}}{3}\)

Consider solving the following equation:

\((x+2)^{2}=25\)

To solve this equation by factoring, first square x+2 and then put it in standard form, equal to zero, by subtracting 25 from both sides.

Factor and then apply the zero-product property.

\(\begin{array}{c}{x^{2}+4 x-21=0} \\ {(x+7)(x-3)=0}\end{array}\)

\(\begin{array}{rr}{x+7=0} & {\text { or } \quad x-3=0} \\ {x=-7} & {x=3}\end{array}\)

The two solutions are -7 and 3.

When an equation is in this form, we can obtain the solutions in fewer steps by extracting the roots.

Example \(\PageIndex{6}\)

\((x=2)^{2}=25\).

Solve by extracting the roots.

\(\begin{array}{rlrr}{(x+2)^{2}} & {=25} & {\color{Cerulean}{ Apply\: the\:square \:root \:property.}} \\ {x+2} & {=\pm \sqrt{25}} & {\color{Cerulean}{ Simplify. }} \\ {x+2}& {=\pm5} \\ {x}&{=-2\pm5}\end{array}\)

At this point, separate the “plus or minus” into two equations and simplify each individually.

\(\begin{array}{ll}{x=-2-5} & {\text { or } x=-2+5} \\ {x=-7} & {\quad\:\: x=3}\end{array}\)

The solutions are -7 and 3.

In addition to fewer steps, this method allows us to solve equations that do not factor.

Example \(\PageIndex{7}\)

\((3 x+3)^{2}-27=0\)

Next, extract the roots and simplify.

\(\begin{aligned}(3 x+3)^{2} &=27 \\ 3 x+3 &=\pm \sqrt{27} \\ 3 x+3 &=\pm \sqrt{9 \cdot 3} \\ 3 x+3 &=\pm 3 \sqrt{3} \end{aligned}\)

Solve for x .

The solutions are \(-1 - \sqrt{3}\) and \(-1 + \sqrt{3}\)

Example \(\PageIndex{8}\)

\(9(2 x-1)^{2}-8=0\)

Begin by isolating the square factor.

Apply the square root property and solve.

The solutions are \(\frac{3 - 2 \sqrt{2}}{6}\) and \(\frac{3 + 2 \sqrt{2}}{6}\)

Exercise \(\PageIndex{2}\)

\(3(x-5)^{2}-2=0\)

\(15\pm 6\sqrt{3}\)

Example \(\PageIndex{9}\)

The length of a rectangle is twice its width. If the diagonal measures 2 feet, then find the dimensions of the rectangle.

Let w represent the width.

Let 2w represent the length.

The diagonal of any rectangle forms two right triangles. Thus the Pythagorean theorem applies. The sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse:

\(\begin{array}{l}{(\operatorname{leg})^{2}+(\operatorname{leg})^{2}=\text { hypotenuse }^{2}} \\ {(\color{OliveGreen}{2 w}\color{black}{)}^{2}+(\color{OliveGreen}{w}\color{black}{)}^{2}=(\color{OliveGreen}{2}\color{black}{)}^{2}}\end{array}\)

\(\begin{aligned}(2 w)^{2}+(w)^{2} &=(2)^{2} \\ 4 w^{2}+w^{2} &=4 \\ 5 w^{2} &=4 \quad\quad\quad\color{Cerulean}{Isolate\:w^{2}.} \\\frac{5w^{2}}{\color{Cerulean}{5}}&\color{black}{=}\frac{4}{\color{Cerulean}{5}}\\ w^{2}&=\frac{4}{5}\quad\:\:\quad\color{Cerulean}{Extract\:the\:roots.} \\ w&=\pm\sqrt{\frac{4}{5}}\\w&=\pm\frac{2}{\sqrt{5}} \end{aligned}\)

Here we obtain two solutions, \(w=−\frac{2}{\sqrt{5}}\) and \(w=\frac{2}{\sqrt{5}}\). Since the problem asked for a length of a rectangle, we disregard the negative answer. Furthermore, we will rationalize the denominator and present our solutions without any radicals in the denominator.

\(\begin{aligned} w &=\frac{2}{\sqrt{5}} \quad\quad\quad\quad\color{Cerulean}{Rationalize\:the\:denominator.} \\ &=\frac{2}{\sqrt{5}} \cdot \color{Cerulean}{\frac{\sqrt{5}}{\sqrt{5}}} \\ &=\frac{2 \sqrt{5}}{\sqrt{25}} \\ &=\frac{2 \sqrt{5}}{5} \end{aligned}\)

Back substitute to find the length.

\(\begin{aligned} l &=2 w \\ &=2\color{black}{\left(\color{OliveGreen}{\frac{2 \sqrt{5}}{5}}\right)} \\ &=\frac{4 \sqrt{5}}{5} \end{aligned}\)

The length of the rectangle is \(\frac{4 \sqrt{5}}{5}\) feet and the width is \(\frac{2 \sqrt{5}}{5}\) feet.

Key Takeaways

• Solve equations of the form \(ax^{2}+c=0\) by extracting the roots.
• Extracting roots involves isolating the square and then applying the square root property. After applying the square root property, you have two linear equations that each can be solved. Be sure to simplify all radical expressions and rationalize the denominator if necessary.

Exercise \(\PageIndex{3}\) extracting square roots

Solve by factoring and then solve by extracting roots. Check answers.

• \(x^{2}-36=0\)
• \(x^{2}-81=0\)
• \(4y^{2}−9=0\)
• \(9y^{2}−25=0\)
• \((x−2)^{2}−1=0\)
• \((x+1)^{2}−4=0\)
• \(4(y−2)^{2}−9=0\)
• \(9(y+1)^{2}−4=0\)
• \(−3(t−1)^{2}+12=0\)
• \(−2(t+1)^{2}+8=0\)
• \((x−5)^{2}−25=0\)
• \((x+2)^{2}−4=0\)

1. \(-6, 6\)

3. \(-\frac{3}{2} , \frac{3}{2}\)

5. \(1, 3\)

7. \(\frac{1}{2}, \frac{7}{2}\)

9. \(-1, 3\)

11. \(0, 10\)

Exercise \(\PageIndex{4}\) extracting square roots

• \(x^{2}=16\)
• \(x^{2}=1\)
• \(y^{2}=9\)
• \(y^{2}=64\)
• \(x^{2}=14\)
• \(x^{2}=19\)
• \(y^{2}=0.25\)
• \(y^{2}=0.04\)
• \(x^{2}=12\)
• \(x^{2}=18\)
• \(16x^{2}=9\)
• \(4x^{2}=25\)
• \(2t^{2} = 1\)
• \(3t^{2} = 2\)
• \(x^{2} −100 = 0\)
• \(x^{2} −121 = 0\)
• \(y^{2} + 4 = 0\)
• \(y^{2} + 1 = 0\)
• \(x^{2} −49 = 0\)
• \(x^{2} −925 = 0\)
• \(y^{2} −0.09 = 0\)
• \(y^{2} −0.81 = 0\)
• \(x^{2} − 7 = 0\)
• \(x^{2} − 2 = 0\)
• \(x^{2} − 8 = 0\)
• \(t^{2} −18 = 0\)
• \(x^{2} + 8 = 0\)
• \(x^{2} +125 = 0\)
• \(16x^{2} −27 = 0\)
• \(9x^{2} − 8 = 0\)
• \(2y^{2} − 3 = 0\)
• \(5y^{2} − 2 = 0\)
• \(3x^{2} − 1 = 0\)
• \(6x^{2} − 3 = 0\)
• \((x + 7 )^{2} − 4 = 0\)
• \((x + 9 )^{2} −36 = 0\)
• \(( 2y − 3 )^{2} −81 = 0 \)
• \(( 2y + 1 )^{2} −25 = 0\)
• \((x − 5 )^{2} −20 = 0\)
• \((x + 1 )^{2} −28 = 0\)
• \(( 3t + 2 )^{2} − 6 = 0\)
• \((3t−5)^{2}−10=0\)
• \(4(y+2)^{2}−3=0\)
• \(9(y−7)^{2}−5=0\)
• \(4(3x+1)^{2}−27=0\)
• \(9(2x−3)^{2}−8=0\)
• \(2(3x−1)^{2}+3=0\)
• \(5(2x−1)^{2}−3=0\)
• \(3(y−23)^{2}−32=0\)
• \(2(3y−13)^{2}−52=0\)

1. ±4

3. ±3

5. ±\(\frac{1}{2}\)

7. ±0.5

9. ±\(2\sqrt{3}\)

11. ±\(\frac{3}{4}\)

13. ±\(\sqrt{\frac{1}{2}}\)

15. ±10

17. No real solution

19. ±\(\frac{2}{3}\)

21. ±0.3

23. ±\(\sqrt{7}\)

25. ±\(2\sqrt{2}\)

27. No real solution

29. ±\(\frac{3 \sqrt{3}}{4}\)

31. ±\(6\sqrt{2}\)

33. ±\(3\sqrt{3}\)

35. \(−9, −5\)

37. \(−3, 6\)

39. \(5\)±\(2\sqrt{5}\)

41. ±\(\frac{\sqrt{6}-2}{3}\)

43. ±\(\frac{\sqrt{3}}{2}-2\)

45. ±\(\frac{3 \sqrt{3}-2}{6}\)

47. No real solution

49. ±\(\frac{4 \sqrt{6}}{3}+23\)

Exercise \(\PageIndex{5}\) extracting square roots

Find a quadratic equation in standard form with the following solutions.

• ±\(\sqrt{7}\)
• ±\(\sqrt{3}\)
• ±\(3\sqrt{5}\)
• ±\(5\sqrt{2}\)
• \(1\)±\(\sqrt{2}\)
• \(2\)±\(\sqrt{3}\)

1. \(x^{2}−49=0\)

3. \(x^{2}−7=0\)

5. \(x^{2}−45=0\)

7. \(x^{2}−2x−1=0\)

Exercise \(\PageIndex{6}\) extracting square roots

Solve and round off the solutions to the nearest hundredth.

• \(9x(x+2)=18x+1\)
• \(x^{2}=10(x^{2}−2)−5\)
• \((x+3)(x−7)=11−4x\)
• \((x−4)(x−3)=66−7x\)
• \((x−2)^{2}=67−4x\)
• \((x+3)^{2}=6x+59\)
• \((2x+1)(x+3)−(x+7)=(x+3)^{2}\)
• \((3x−1)(x+4)=2x(x+6)−(x−3)\)

1. ±0.33

3. ±5.66

5. ±7.94

7. ±3.61

Exercise \(\PageIndex{7}\) extracting square roots

Set up an algebraic equation and use it to solve the following.

• If 9 is subtracted from 4 times the square of a number, then the result is 3. Find the number.
• If 20 is subtracted from the square of a number, then the result is 4. Find the number.
• If 1 is added to 3 times the square of a number, then the result is 2. Find the number.
• If 3 is added to 2 times the square of a number, then the result is 12. Find the number.
• If a square has an area of 8 square centimeters, then find the length of each side.
• If a circle has an area of 32\(\pi\) square centimeters, then find the length of the radius.
• The volume of a right circular cone is 36\(\pi\) cubic centimeters when the height is 6 centimeters. Find the radius of the cone. (The volume of a right circular cone is given by \(V=13\pi r^{2}h\).)
• The surface area of a sphere is 75\(\pi\) square centimeters. Find the radius of the sphere. (The surface area of a sphere is given by SA=\(4\pi r^{2}\).)
• The length of a rectangle is 6 times its width. If the area is 96 square inches, then find the dimensions of the rectangle.
• The base of a triangle is twice its height. If the area is 16 square centimeters, then find the length of its base.
• A square has an area of 36 square units. By what equal amount will the sides have to be increased to create a square with double the given area?
• A circle has an area of 25\(\pi\) square units. By what amount will the radius have to be increased to create a circle with double the given area?
• If the sides of a square measure 1 unit, then find the length of the diagonal.
• If the sides of a square measure 2 units, then find the length of the diagonal.
• The diagonal of a square measures 5 inches. Find the length of each side.
• The diagonal of a square measures 3 inches. Find the length of each side.
• The length of a rectangle is twice its width. If the diagonal measures 10 feet, then find the dimensions of the rectangle.
• The length of a rectangle is twice its width. If the diagonal measures 8 feet, then find the dimensions of the rectangle.
• The length of a rectangle is 3 times its width. If the diagonal measures 5 meters, then find the dimensions of the rectangle.
• The length of a rectangle is 3 times its width. If the diagonal measures 2 feet, then find the dimensions of the rectangle.
• The height in feet of an object dropped from a 9-foot ladder is given by h(t)\(=−16t^{2}+9\), where t represents the time in seconds after the object has been dropped. How long does it take the object to hit the ground? (Hint: The height is 0 when the object hits the ground.)
• The height in feet of an object dropped from a 20-foot platform is given by h(t)\(=−16t^{2}+20\), where t represents the time in seconds after the object has been dropped. How long does it take the object to hit the ground?
• How long will it take to reach half of the distance to the ground, 72 feet?
• How long will it take to travel the rest of the distance to the ground? Round off to the nearest hundredth of a second.
• How long will it take to reach half of the distance to the ground?
• Create an equation of your own that can be solved by extracting the root. Share it, along with the solution, on the discussion board.
• Explain why the technique of extracting roots greatly expands our ability to solve quadratic equations.
• Explain in your own words how to solve by extracting the roots. 106. Derive a formula for the diagonal of a square in terms of its sides.

1. \(−\sqrt{3}\) or \(\sqrt{3}\)

3. \(−3\sqrt{3}\) or \(3\sqrt{3}\)

5. \(2\sqrt{2}\) centimeters

7. \(3\sqrt{2}\) centimeters

9. Length: 24 inches; width: 4 inches

11. \(−6+6\sqrt{2}\approx 2.49\) units

13. \(\sqrt{2}\) units

15. \(52\sqrt{2}\) inches

17. Length: \(4\sqrt{5}\) feet; width: \(2\sqrt{5}\) feet

19. Length: \(310−−\sqrt{2}\) meters; width: \(10−−\sqrt{2}\) meters

21. \(\frac{3}{4}\) second

23. a. 2.12 seconds b. 0.88 second

25. Answers may vary

A Step-by-Step Guide to A3 Problem Solving Methodology

Author: Daniel Croft

Daniel Croft is an experienced continuous improvement manager with a Lean Six Sigma Black Belt and a Bachelor's degree in Business Management. With more than ten years of experience applying his skills across various industries, Daniel specializes in optimizing processes and improving efficiency. His approach combines practical experience with a deep understanding of business fundamentals to drive meaningful change.

Problem-solving is an important component of any business or organization. It entails identifying, analyzing, and resolving problems in order to improve processes, drive results, and foster a culture of continuous improvement. A3 Problem solving is one of the most effective problem-solving methodologies.

A3 Problem solving is a structured and systematic approach to problem-solving that originated with the lean manufacturing methodology. It visualizes the problem-solving process using a one-page document known as an A3 report. The A3 report provides an overview of the problem, data analysis, root causes, solutions, and results in a clear and concise manner.

A3 Problem Solving has numerous advantages, including improved communication, better decision-making, increased efficiency, and reduced waste. It is a powerful tool for businesses of all sizes and industries, and it is especially useful for solving complex and multi-faceted problems.

In this blog post, we will walk you through the A3 Problem Solving methodology step by step. Whether you are new to A3 Problem Solving or simply want to improve your skills, this guide will help you understand and apply the process in your workplace.

What is A3 Problem Solving?

A3 Problem Solving is a structured and systematic approach to problem-solving that makes use of a one-page document called an A3 report to visually represent the process. The A3 report provides an overview of the problem, data analysis, root causes, solutions, and results in a clear and concise manner. The method was created within the framework of the Lean manufacturing methodology and is based on the principles of continuous improvement and visual management.

Looking for a A3 Problem solving template? Click here

Origin and History of A3 Problem Solving

A3 Problem Solving was developed by Toyota Motor Corporation and was first used in the manufacture of automobiles. The term “A3” refers to the size of the paper used to create the report, which is an ISO standard known as “A3”. The goal of the A3 report is to provide a visual representation of the problem-solving process that all members of the organisation can easily understand and share. A3 Problem Solving has been adopted by organisations in a variety of industries over the years, and it has become a widely used and recognised method for problem-solving.

Key Principles of A3 Problem Solving

The following are the key principles of A3 Problem Solving:

• Define the problem clearly and concisely
• Gather and analyze data to gain a deep understanding of the problem
• Identify the root causes of the problem
• Develop and implement effective solutions
• Evaluate results and continuously improve

These principles serve as the foundation of the A3 Problem Solving methodology and are intended to assist organisations in continuously improving and achieving their objectives. Organizations can effectively solve problems, identify areas for improvement, and drive results by adhering to these principles.

Step 1: Define the Problem

Importance of clearly defining the problem.

The first step in the A3 Problem Solving process is critical because it lays the groundwork for the remaining steps. To define the problem clearly and accurately, you must first understand the problem and identify the underlying root cause. This step is critical because if the problem is not correctly defined, the rest of the process will be based on incorrect information, and the solution developed may not address the issue effectively.

The significance of defining the problem clearly cannot be overstated. It aids in the collection and analysis of relevant data, which is critical for developing effective solutions. When the problem is clearly defined, the data gathered is more relevant and targeted, resulting in a more comprehensive understanding of the issue. This will enable the development of solutions that are more likely to be effective because they are founded on a thorough and accurate understanding of the problem.

However, if the problem is not clearly defined, the data gathered may be irrelevant or incorrect, resulting in incorrect conclusions and ineffective solutions. Furthermore, the process of collecting and analysing data can become time-consuming and inefficient, resulting in resource waste. Furthermore, if the problem is not accurately defined, the solutions developed may fail to address the root cause of the problem, resulting in ongoing issues and a lack of improvement.

Techniques for Defining the Problem

The first step in the A3 Problem Solving process is to clearly and accurately define the problem. This is an important step because a clearly defined problem will help to ensure that the appropriate data is collected and solutions are developed. If the problem is not clearly defined, incorrect data may be collected, solutions that do not address the root cause of the problem, and time and resources may be wasted.

A problem can be defined using a variety of techniques, including brainstorming , root cause analysis , process mapping , and Ishikawa diagrams . Each of these techniques has its own advantages and disadvantages and can be used in a variety of situations depending on the nature of the problem.

Best Practice for Defining the Problem

In addition to brainstorming, root cause analysis, process mapping, and Ishikawa diagram s, best practices should be followed when defining a problem in A3 Problem Solving. Among these best practices are:

• Define the issue in a specific and quantifiable way: It is critical to be specific and concise when defining the problem, as well as to quantify the problem in terms of its impact. This will help to ensure that all stakeholders understand the problem and that data collection is focused on the right areas.
• Focus on the problem’s root cause: The A3 Problem Solving methodology is intended to assist organisations in identifying and addressing the root cause of a problem, rather than just the symptoms. Organizations can ensure that their solutions are effective and long-lasting by focusing on the root cause of the problem.
• Ascertain that all stakeholders agree on the problem’s definition: All stakeholders must agree on the definition of the problem for the A3 Problem Solving process to be effective. This ensures that everyone is working towards the same goal and that the solutions developed are relevant and appropriate.
• Consider the problem’s impact on the organisation and its stakeholders: It is critical to consider the impact of the problem on the organisation and its stakeholders when defining it. This will assist in ensuring that the appropriate data is gathered and that the solutions developed are relevant and appropriate.

Organizations can ensure that their problem is defined in a way that allows for effective data collection, analysis, and solution development by following these best practices. This will aid in the development of appropriate solutions and the effective resolution of the problem, resulting in improvements in the organization’s processes and outcomes.

Step 2: Gather Data

Gathering data in a3 problem solving.

Data collection is an important step in the A3 Problem Solving process because it allows organisations to gain a thorough understanding of the problem they are attempting to solve. This step entails gathering pertinent information about the problem, such as data on its origin, impact, and any related factors. This information is then used to help identify root causes and develop effective solutions.

One of the most important advantages of data collection in A3 Problem Solving is that it allows organisations to identify patterns and trends in data, which can be useful in determining the root cause of the problem. This information can then be used to create effective solutions that address the problem’s root cause rather than just its symptoms.

In A3 Problem Solving, data collection is a collaborative effort involving all stakeholders, including those directly impacted by the problem and those with relevant expertise or experience. Stakeholders can ensure that all relevant information is collected and that the data is accurate and complete by working together.

Overall, data collection is an important step in the A3 Problem Solving process because it serves as the foundation for effective problem-solving. Organizations can gain a deep understanding of the problem they are attempting to solve and develop effective solutions that address its root cause by collecting and analysing relevant data.

Data Collection Methods

In A3 Problem Solving, several data collection methods are available, including:

• Observations
• Process diagrams

The best data collection method will be determined by the problem being solved and the type of data required. To gain a complete understanding of the problem, it is critical to use multiple data collection methods.

Tools for Data Analysis and Visualization

Once the data has been collected, it must be analysed and visualised in order to gain insights into the problem. This process can be aided by the following tools:

• Excel Spreadsheets
• Flow diagrams
• Pareto diagrams
• Scatter Plots
• Control diagrams

These tools can assist in organising data and making it easier to understand. They can also be used to generate visual representations of data, such as graphs and charts, to communicate the findings to others.

Finally, the data collection and analysis step is an important part of the A3 Problem Solving process. Organizations can gain a better understanding of the problem and develop effective solutions by collecting and analysing relevant data.

Step 3: Identify Root Causes

Identifying the root causes of the problem is the third step in the A3 Problem Solving process. This step is critical because it assists organisations in understanding the root causes of a problem rather than just its symptoms. Once the underlying cause of the problem is identified, it can be addressed more effectively, leading to more long-term solutions.

Overview of the Root Cause Analysis Process

The process of determining the underlying causes of a problem is known as root cause analysis. This process can assist organisations in determining why a problem is occurring and what can be done to prevent it from recurring in the future. The goal of root cause analysis is to identify the underlying cause of a problem rather than just its symptoms, allowing it to be addressed more effectively.

To understand Root cause analysis in more detail check out RCA in our Lean Six Sigma Yellow Belt Course Root Cause Analysis section

Techniques for Identifying Root Causes

There are several techniques for determining the root causes of a problem, including:

• Brainstorming
• Ishikawa diagrams (also known as fishbone diagrams)
• Root Cause Tree Analysis

These methods can be used to investigate the issue in-depth and identify potential root causes. Organizations can gain a deeper understanding of the problem and identify the underlying causes that must be addressed by using these techniques.

Best Practices for Conducting Root Cause Analysis

It is critical to follow these best practices when conducting root cause analysis in A3 Problem Solving:

• Make certain that all stakeholders participate in the root cause analysis process.
• Concentrate on determining the root cause of the problem rather than just its symptoms.
• Take into account all potential root causes, not just the most obvious ones.
• To identify root causes, use a systematic approach, such as the 5 Whys or root cause tree analysis.

Organizations can ensure that root cause analysis is carried out effectively and that the root cause of the problem is identified by adhering to these best practises. This will aid in the development of appropriate solutions and the effective resolution of the problem.

Step 4: Develop Solutions

Developing solutions is the fourth step in the A3 Problem Solving process. This entails generating ideas and options for dealing with the problem, followed by selecting the best solution. The goal is to develop a solution that addresses the root cause of the problem and prevents it from recurring.

Solution Development in A3 Problem Solving

A3 solution development Problem solving is an iterative process in which options are generated and evaluated. The data gathered in the previous steps, as well as the insights and understanding gained from the root cause analysis, guide this process. The solution should be based on a thorough understanding of the problem and address the underlying cause.

Techniques for Developing Solutions

There are several techniques that can be used to develop solutions in A3 Problem Solving, including:

• Brainwriting
• Solution matrix
• Multi voting
• Force field analysis

These techniques can help to generate a range of options and to select the best solution.

Best Practice for Developing Solutions

It is critical to follow the following best practices when developing solutions in A3 Problem Solving:

• Participate in the solution development process with all stakeholders.
• Make certain that the solution addresses the underlying cause of the problem.
• Make certain that the solution is feasible and achievable.
• Consider the solution’s impact on the organisation and its stakeholders.

Organizations can ensure that the solutions they develop are effective and sustainable by adhering to these best practises. This will help to ensure that the problem is addressed effectively and that it does not reoccur.

Step 5: Implement Solutions

The final and most important step in the A3 Problem Solving methodology is solution implementation. This is the stage at which the identified and developed solutions are put into action to address the problem. This step’s goal is to ensure that the solutions are effective, efficient, and long-lasting.

The implementation Process

The implementation process entails putting the solutions developed in the previous step into action. This could include changes to processes, procedures, and systems, as well as employee training and education. To ensure that the solutions are effective, the implementation process should be well-planned and meticulously executed.

Techniques for Implementing Solutions

A3 Problem Solving solutions can be implemented using a variety of techniques, including:

• Piloting the solution on a small scale before broadening its application
• Participating in the implementation process with all relevant stakeholders
• ensuring that the solution is in line with the goals and objectives of the organisation
• Monitoring the solution to determine its effectiveness and make any necessary changes

Best Practice for Implementing Solutions

It is critical to follow these best practices when implementing solutions in A3 Problem Solving:

Make certain that all relevant stakeholders are involved and supportive of the solution. Have a clear implementation plan that outlines the steps, timeline, and resources required. Continuously monitor and evaluate the solution to determine its efficacy and make any necessary changes. Encourage all stakeholders to communicate and collaborate openly. Organizations can ensure that solutions are effectively implemented and problems are effectively addressed by adhering to these best practices. The ultimate goal is to find a long-term solution to the problem and improve the organization’s overall performance.

In conclusion, A3 Problem Solving is a comprehensive and structured methodology for problem-solving that can be applied in various industries and organisations. The A3 Problem Solving process’s five steps – Define the Problem, Gather Data, Identify Root Causes, Develop Solutions, and Implement Solutions – provide a road map for effectively addressing problems and making long-term improvements.

Organizations can improve their problem-solving skills and achieve better results by following the key principles, techniques, and best practices outlined in this guide. As a result, both the organisation and its stakeholders will benefit from increased efficiency, effectiveness, and satisfaction. So, whether you’re an experienced problem solver or just getting started, consider incorporating the A3 Problem Solving methodology into your work and start reaping the benefits right away.

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Daniel Croft

Hi im Daniel continuous improvement manager with a Black Belt in Lean Six Sigma and over 10 years of real-world experience across a range sectors, I have a passion for optimizing processes and creating a culture of efficiency. I wanted to create Learn Lean Siigma to be a platform dedicated to Lean Six Sigma and process improvement insights and provide all the guides, tools, techniques and templates I looked for in one place as someone new to the world of Lean Six Sigma and Continuous improvement.

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• Square Roots

Square Root Of 3

The  square root of 3   is represented using the square root or the radical symbol “ √”, and it is written as  √3. The value of  √3 is approximately equal to  1.732 . This value is widely used in mathematics. Since root 3 is an  irrational number , which cannot be represented in the form of a fraction. It means that it has an infinite number of decimals. Here, we are going to have a look at the value of root and the long division method to find the value of root 3.

Square Root of 3 Value

The square root of 3 is represented as  √3  or 3 1/2 . The numerical value of root 3 is given as follows:

√3 = 1.732050807568877293527446341505872366….. 

To make the calculation easier, the value of root 3 is truncated as 1.732. 

How to Find the Square Root of 3?

Generally, before finding the value for the square root of a number, it is necessary to check whether the given number is a perfect square or not. Because we can easily find the square root values for perfect squares. But, finding the square root value for non-perfect squares, is quite difficult. So, we use the long division method to find the value of the square root of 3.

Square Root of 3 by Long Division Method

Here, we use the long division method to obtain the value of the square root of 3.

Thus, the value of the square root of 3 is 1.732.

Video Lessons on Square Roots

Visualising square roots.

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Screen Rant

The wheel of time season 3 can solve 1 egwene problem with a time skip.

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Wheel Of Time Has Set Up 3 Huge Egwene Stories From The Books

Wheel of time's show hints rand will drop a huge part of his strategy from the books, the wheel of time season 2 confirmed a mat detail robert jordan debunked 22 years ago.

The next season of Amazon Prime Video's TV show adaptation of The Wheel of Time will focus on a new book in the famed fantasy series by Robert Jordan and Brandon Sanderson. Madeleine Madden has played Egwene al'Vere with fierce determination since the beginning of the series, and The Wheel of Time must shift its trajectory to do her character justice. As Egwene is a huge part of the show and novels but hasn't been Accepted yet, which would mean she couldn't join the other characters in their arcs in season 3. However, a time jump could fix this.

After the intense conclusion of season 2, there are theories that The Wheel of Time season 3 will cover several books , which would be beneficial for the plot. If The Wheel of Time skips book 3, The Dragon Reborn, and only focuses on book 4, The Shadow Rising , it would spell trouble for certain character arcs, including Egwene's. A time-skip would ease the transition from The Great Hunt story to The Shadow Rising. Additionally, it would also allow the characters to develop their powers and abilities off-screen.

The Wheel Of Time Has Given Itself An Egwene Problem Heading Into Season 3

Egwene still needs more training before she can go to the aiel waste.

The Wheel of Time season 2 ending was an intense episode with many character deaths and significant changes to the story. Egwene left off the story as a novice, but this will be a challenge for the story set up by the finale. The relationship between Egwene and Rand is a significant part of the story, and when they're reunited in the Aiel Waste in The Shadow Rising , the pair have to reckon with their feelings for each other. However, this is a dangerous journey, and without the additional time she gets in The Dragon Reborn , Egwene wouldn't be ready.

Egwene hasn't progressed enough in their Aes Sedai training to justify their traveling to the Waste with the Aiel and Rand, but splitting them up would halt the momentum from season 2. The Wheel of Time season 3 needs Egwene to be Accepted off-screen, even if this deviates from the books. If this doesn't happen, Egwene would have to go back to the White Tower and be absent from the trip to the Aiel Waste, or the Waste storyline would have to be postponed entirely.

Nynaeve has already passed her Accepted test in The Wheel of Time show.

Egwene al'Vere is one of The Wheel of Time's main characters, and the TV show is setting the gears in motion for Egwene to follow her book story.

A Time Skip Could Help Adapt Egwene's Story Properly In The Wheel Of Time TV Show

Egwene will pass the accepted test and be ready to join the other characters.

Allowing some of the events of The Dragon Reborn , like Egwene recovering from her time with the Seanchan and becoming Accepted, to happen during a time jump would be a great solution for The Wheel of Time . The series has already spent many episodes establishing the lore and importance of the Aiel's paths to power, and Egwene has a bright future ahead of her, according to the books. Trusting the audience to remember what happened in season 2 and recall how Egwene had great potential would let the show move forward without wasting additional time.

It won't take much convincing for audiences to acknowledge that the young novice Egwene used to be has transformed into someone with the capacity for Dreaming.

Though she, Nynaeve, and Elayne needed additional seasoning before being sold to the Seanchan, they all have the capacity for power and ability. It won't take much convincing for audiences to acknowledge that the young novice Egwene used to be has transformed into someone with the capacity for Dreaming. In order to preserve Egwene's story in The Wheel of Time season 3 , the series can allude to her return to the White Tower. An acceleration in the timeline would also account for the years between the release of The Wheel of Time season 2 and season 3.

Season 3 Can't Completely Ignore The Wise Ones' Teachings & Accepted Test

Commitment to the established worldbuilding will be critical for the wheel of time.

Even if The Wheel of Time does include a time jump, that doesn't mean it should overlook the previously established lore and story beats of The Dragon Reborn . It would be a mistake for The Wheel of Time to jump into season 3 exactly where the narrative left off in the finale of season 2 and have Egwene go to the Aiel Waste as she currently is in the show. The series has already demonstrated how intensive and critical the Accepted test is to the experience of novitiates, and going back on this would undermine the story's strength.

Moving forward without allowing Egwene to learn more at the White Tower and be Accepted would throw her Dreaming narrative off course and change the events of The Shadow Rising . Egwene will play a vital role in The Wheel of Time season 3, and she deserves time to mature to handle the responsibility she's about to assume. Additionally, traveling with Rand again while undergoing these changes is something an older and wiser Egwene undergoes in the books, and the same should be true in the series.

The Wheel of Time

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Brandon Sanderson and Robert Jordan's expansive fantasy series is brought to life in The Wheel of Time, a fantasy tv series created for Amazon Prime Video. The series follows a woman named Moiraine, a member of the all-female Aes Sedai group who can utilize a great power. Following an attack on a local village, Moiraine heads there to find a villager that may be the reincarnation of an all-powerful dragon that will either save or destroy the world.

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• To find the roots factor the function, set each facotor to zero, and solve. The solutions are the roots of the function.
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• A root is a value for which the function equals zero. The roots are the points where the function intercept with the x-axis
• What are complex roots?
• Complex roots are the imaginary roots of a function.
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• To find the complex roots of a quadratic equation use the formula: x = (-b±i√(4ac – b2))/2a

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Title: fast analysis of the openai o1-preview model in solving random k-sat problem: does the llm solve the problem itself or call an external sat solver.

Abstract: In this manuscript I present an analysis on the performance of OpenAI O1-preview model in solving random K-SAT instances for K$\in {2,3,4}$ as a function of $\alpha=M/N$ where $M$ is the number of clauses and $N$ is the number of variables of the satisfiable problem. I show that the model can call an external SAT solver to solve the instances, rather than solving them directly. Despite using external solvers, the model reports incorrect assignments as output. Moreover, I propose and present an analysis to quantify whether the OpenAI O1-preview model demonstrates a spark of intelligence or merely makes random guesses when outputting an assignment for a Boolean satisfiability problem.

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