problem solving square root

• PRO Courses Guides New Tech Help Pro Expert Videos About wikiHow Pro Upgrade Sign In
• EDIT Edit this Article
• EXPLORE Tech Help Pro About Us Random Article Quizzes Request a New Article Community Dashboard This Or That Game Forums Popular Categories Arts and Entertainment Artwork Books Movies Computers and Electronics Computers Phone Skills Technology Hacks Health Men's Health Mental Health Women's Health Relationships Dating Love Relationship Issues Hobbies and Crafts Crafts Drawing Games Education & Communication Communication Skills Personal Development Studying Personal Care and Style Fashion Hair Care Personal Hygiene Youth Personal Care School Stuff Dating All Categories Arts and Entertainment Finance and Business Home and Garden Relationship Quizzes Cars & Other Vehicles Food and Entertaining Personal Care and Style Sports and Fitness Computers and Electronics Health Pets and Animals Travel Education & Communication Hobbies and Crafts Philosophy and Religion Work World Family Life Holidays and Traditions Relationships Youth
• Browse Articles
• Learn Something New
• Quizzes Hot
• Happiness Hub
• This Or That Game
• Train Your Brain
• Explore More
• Support wikiHow
• About wikiHow
• Log in / Sign up
• Education and Communications
• Mathematics

How to Solve Square Root Problems

Last Updated: October 22, 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 12 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 410,125 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. [21] X Research source 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 . 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
• ↑ https://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)
• ↑ https://mathbitsnotebook.com/Algebra1/Radicals/RADNegativeUnder.html
• ↑ https://nrich.maths.org/problems/unusual-long-division-square-roots-calculators
• ↑ https://www.cuemath.com/algebra/square-root-by-long-division-method/
• ↑ https://www.calculator.net/root-calculator.html

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

• Send fan mail to authors

Reader Success Stories

Laura Richards

May 21, 2017

Did this article help you?

Matthew LeCounte

Sep 17, 2020

May 5, 2016

Featured Articles

Trending Articles

Watch Articles

• Terms of Use
• Privacy Policy
• Do Not Sell or Share My Info
• Not Selling Info

wikiHow Tech Help Pro:

Develop the tech skills you need for work and life

High Impact Tutoring Built By Math Experts

Personalized standards-aligned one-on-one math tutoring for schools and districts

In order to access this I need to be confident with:

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

Still stuck?

At Third Space Learning, we specialize in helping teachers and school leaders to provide personalized math support for more of their students through high-quality, online one-on-one math tutoring delivered by subject experts.

Each week, our tutors support thousands of students who are at risk of not meeting their grade-level expectations, and help accelerate their progress and boost their confidence.

Find out how we can help your students achieve success with our math tutoring programs .

[FREE] Common Core Practice Tests (3rd to 8th Grade)

Prepare for math tests in your state with these 3rd Grade to 8th Grade practice assessments for Common Core and state equivalents.

Get your 6 multiple choice practice tests with detailed answers to support test prep, created by US math teachers for US math teachers!

Privacy Overview

9.6 Solve Equations with Square Roots

Learning objectives.

By the end of this section, you will be able to:

• Solve radical equations
• Use square roots in applications

Be Prepared 9.15

Before you get started, take this readiness quiz.

Simplify: ⓐ 9 9 ⓑ 9 2 9 2 . If you missed this problem, review Example 9.1 and Example 1.19 .

Be Prepared 9.16

Solve: 5 ( x + 1 ) − 4 = 3 ( 2 x − 7 ) 5 ( x + 1 ) − 4 = 3 ( 2 x − 7 ) . If you missed this problem, review Example 2.42 .

Be Prepared 9.17

Solve: n 2 − 6 n + 8 = 0 n 2 − 6 n + 8 = 0 . If you missed this problem, review Example 7.73 .

Solve Radical Equations

In this section we will solve equations that have the variable in the radicand of a square root. Equations of this type are called radical equations.

Radical Equation

An equation in which the variable is in the radicand of a square root is called a radical equation .

As usual, in solving these equations, what we do to one side of an equation we must do to the other side as well. Since squaring a quantity and taking a square root are ‘opposite’ operations, we will square both sides in order to remove the radical sign and solve for the variable inside.

But remember that when we write a a we mean the principal square root. So a ≥ 0 a ≥ 0 always. When we solve radical equations by squaring both sides we may get an algebraic solution that would make a a negative. This algebraic solution would not be a solution to the original radical equation ; it is an extraneous solution. We saw extraneous solutions when we solved rational equations, too.

Example 9.74

For the equation x + 2 = x x + 2 = x :

ⓐ Is x = 2 x = 2 a solution? ⓑ Is x = −1 x = −1 a solution?

ⓐ Is x = 2 x = 2 a solution?

ⓑ Is x = −1 x = −1 a solution?

Try It 9.147

For the equation x + 6 = x x + 6 = x :

ⓐ Is x = −2 x = −2 a solution? ⓑ Is x = 3 x = 3 a solution?

Try It 9.148

For the equation − x + 2 = x − x + 2 = x :

ⓐ Is x = −2 x = −2 a solution? ⓑ Is x = 1 x = 1 a solution?

Now we will see how to solve a radical equation. Our strategy is based on the relation between taking a square root and squaring.

Example 9.75

How to solve radical equations.

Solve: 2 x − 1 = 7 2 x − 1 = 7 .

Try It 9.149

Solve: 3 x − 5 = 5 3 x − 5 = 5 .

Try It 9.150

Solve: 4 x + 8 = 6 4 x + 8 = 6 .

Solve a radical equation.

• Step 1. Isolate the radical on one side of the equation.
• Step 2. Square both sides of the equation.
• Step 3. Solve the new equation.
• Step 4. Check the answer.

Example 9.76

Solve: 5 n − 4 − 9 = 0 5 n − 4 − 9 = 0 .

Try It 9.151

Solve: 3 m + 2 − 5 = 0 3 m + 2 − 5 = 0 .

Try It 9.152

Solve: 10 z + 1 − 2 = 0 10 z + 1 − 2 = 0 .

Example 9.77

Solve: 3 y + 5 + 2 = 5 3 y + 5 + 2 = 5 .

Try It 9.153

Solve: 3 p + 3 + 3 = 5 3 p + 3 + 3 = 5 .

Try It 9.154

Solve: 5 q + 1 + 4 = 6 5 q + 1 + 4 = 6 .

When we use a radical sign, we mean the principal or positive root. If an equation has a square root equal to a negative number, that equation will have no solution.

Example 9.78

Solve: 9 k − 2 + 1 = 0 9 k − 2 + 1 = 0 .

Try It 9.155

Solve: 2 r − 3 + 5 = 0 2 r − 3 + 5 = 0 .

Try It 9.156

Solve: 7 s − 3 + 2 = 0 7 s − 3 + 2 = 0 .

If one side of the equation is a binomial, we use the binomial squares formula when we square it.

Binomial Squares

Don’t forget the middle term!

Example 9.79

Solve: p − 1 + 1 = p p − 1 + 1 = p .

Try It 9.157

Solve: x − 2 + 2 = x x − 2 + 2 = x .

Try It 9.158

Solve: y − 5 + 5 = y y − 5 + 5 = y .

Example 9.80

Solve: r + 4 − r + 2 = 0 r + 4 − r + 2 = 0 .

Try It 9.159

Solve: m + 9 − m + 3 = 0 m + 9 − m + 3 = 0 .

Try It 9.160

Solve: n + 1 − n + 1 = 0 n + 1 − n + 1 = 0 .

When there is a coefficient in front of the radical, we must square it, too.

Example 9.81

Solve: 3 3 x − 5 − 8 = 4 3 3 x − 5 − 8 = 4 .

Try It 9.161

Solve: 2 4 a + 2 − 16 = 16 2 4 a + 2 − 16 = 16 .

Try It 9.162

Solve: 3 6 b + 3 − 25 = 50 3 6 b + 3 − 25 = 50 .

Example 9.82

Solve: 4 z − 3 = 3 z + 2 4 z − 3 = 3 z + 2 .

Try It 9.163

Solve: 2 x − 5 = 5 x + 3 2 x − 5 = 5 x + 3 .

Try It 9.164

Solve: 7 y + 1 = 2 y − 5 7 y + 1 = 2 y − 5 .

Sometimes after squaring both sides of an equation, we still have a variable inside a radical. When that happens, we repeat Step 1 and Step 2 of our procedure. We isolate the radical and square both sides of the equation again.

Example 9.83

Solve: m + 1 = m + 9 m + 1 = m + 9 .

Try It 9.165

Solve: x + 3 = x + 5 x + 3 = x + 5 .

Try It 9.166

Solve: m + 5 = m + 16 m + 5 = m + 16 .

Example 9.84

Solve: q − 2 + 3 = 4 q + 1 q − 2 + 3 = 4 q + 1 .

Try It 9.167

Solve: y − 3 + 2 = 4 y + 2 y − 3 + 2 = 4 y + 2 .

Try It 9.168

Solve: n − 4 + 5 = 3 n + 3 n − 4 + 5 = 3 n + 3 .

Use Square Roots in Applications

As you progress through your college courses, you’ll encounter formulas that include square roots in many disciplines. We have already used formulas to solve geometry applications.

We will use our Problem Solving Strategy for Geometry Applications, with slight modifications, to give us a plan for solving applications with formulas from any discipline.

Solve applications with formulas.

• Step 1. Read the problem and make sure all the words and ideas are understood. When appropriate, draw a figure and label it with the given information.
• Step 2. Identify what we are looking for.
• Step 3. Name what we are looking for by choosing a variable to represent it.
• Step 4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
• Step 5. Solve the equation using good algebra techniques.
• Step 6. Check the answer in the problem and make sure it makes sense.
• Step 7. Answer the question with a complete sentence.

We used the formula A = L · W A = L · W to find the area of a rectangle with length L and width W . A square is a rectangle in which the length and width are equal. If we let s be the length of a side of a square, the area of the square is s 2 s 2 .

The formula A = s 2 A = s 2 gives us the area of a square if we know the length of a side. What if we want to find the length of a side for a given area? Then we need to solve the equation for s .

A = s 2 Take the square root of both sides. A = s 2 Simplify. A = s A = s 2 Take the square root of both sides. A = s 2 Simplify. A = s

We can use the formula s = A s = A to find the length of a side of a square for a given area.

Area of a Square

We will show an example of this in the next example.

Example 9.85

Mike and Lychelle want to make a square patio. They have enough concrete to pave an area of 200 square feet. Use the formula s = A s = A to find the length of each side of the patio. Round your answer to the nearest tenth of a foot.

Try It 9.169

Katie wants to plant a square lawn in her front yard. She has enough sod to cover an area of 370 square feet. Use the formula s = A s = A to find the length of each side of her lawn. Round your answer to the nearest tenth of a foot.

Try It 9.170

Sergio wants to make a square mosaic as an inlay for a table he is building. He has enough tile to cover an area of 2704 square centimeters. Use the formula s = A s = A to find the length of each side of his mosaic. Round your answer to the nearest tenth of a centimeter.

Another application of square roots has to do with gravity.

Falling Objects

On Earth, if an object is dropped from a height of h h feet, the time in seconds it will take to reach the ground is found by using the formula,

For example, if an object is dropped from a height of 64 feet, we can find the time it takes to reach the ground by substituting h = 64 h = 64 into the formula.

It would take 2 seconds for an object dropped from a height of 64 feet to reach the ground.

Example 9.86

Christy dropped her sunglasses from a bridge 400 feet above a river. Use the formula t = h 4 t = h 4 to find how many seconds it took for the sunglasses to reach the river.

Try It 9.171

A helicopter dropped a rescue package from a height of 1,296 feet. Use the formula t = h 4 t = h 4 to find how many seconds it took for the package to reach the ground.

Try It 9.172

A window washer dropped a squeegee from a platform 196 feet above the sidewalk Use the formula t = h 4 t = h 4 to find how many seconds it took for the squeegee to reach the sidewalk.

Police officers investigating car accidents measure the length of the skid marks on the pavement. Then they use square roots to determine the speed, in miles per hour, a car was going before applying the brakes.

Skid Marks and Speed of a Car

If the length of the skid marks is d feet, then the speed, s , of the car before the brakes were applied can be found by using the formula,

Example 9.87

After a car accident, the skid marks for one car measured 190 feet. Use the formula s = 24 d s = 24 d to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Try It 9.173

An accident investigator measured the skid marks of the car. The length of the skid marks was 76 feet. Use the formula s = 24 d s = 24 d to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Try It 9.174

The skid marks of a vehicle involved in an accident were 122 feet long. Use the formula s = 24 d s = 24 d to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Section 9.6 Exercises

Practice makes perfect.

In the following exercises, check whether the given values are solutions.

For the equation x + 12 = x x + 12 = x : ⓐ Is x = 4 x = 4 a solution? ⓑ Is x = −3 x = −3 a solution?

For the equation − y + 20 = y − y + 20 = y : ⓐ Is y = 4 y = 4 a solution? ⓑ Is y = −5 y = −5 a solution?

For the equation t + 6 = t t + 6 = t : ⓐ Is t = −2 t = −2 a solution? ⓑ Is t = 3 t = 3 a solution?

For the equation u + 42 = u u + 42 = u : ⓐ Is u = −6 u = −6 a solution? ⓑ Is u = 7 u = 7 a solution?

In the following exercises, solve.

5 y + 1 = 4 5 y + 1 = 4

7 z + 15 = 6 7 z + 15 = 6

5 x − 6 = 8 5 x − 6 = 8

4 x − 3 = 7 4 x − 3 = 7

2 m − 3 − 5 = 0 2 m − 3 − 5 = 0

2 n − 1 − 3 = 0 2 n − 1 − 3 = 0

6 v − 2 − 10 = 0 6 v − 2 − 10 = 0

4 u + 2 − 6 = 0 4 u + 2 − 6 = 0

5 q + 3 − 4 = 0 5 q + 3 − 4 = 0

4 m + 2 + 2 = 6 4 m + 2 + 2 = 6

6 n + 1 + 4 = 8 6 n + 1 + 4 = 8

2 u − 3 + 2 = 0 2 u − 3 + 2 = 0

5 v − 2 + 5 = 0 5 v − 2 + 5 = 0

3 z − 5 + 2 = 0 3 z − 5 + 2 = 0

2 m + 1 + 4 = 0 2 m + 1 + 4 = 0

ⓐ u − 3 + 3 = u u − 3 + 3 = u ⓑ x + 1 − x + 1 = 0 x + 1 − x + 1 = 0

ⓐ v − 10 + 10 = v v − 10 + 10 = v ⓑ y + 4 − y + 2 = 0 y + 4 − y + 2 = 0

ⓐ r − 1 − r = −1 r − 1 − r = −1 ⓑ z + 100 − z + 10 = 0 z + 100 − z + 10 = 0

ⓐ s − 8 − s = −8 s − 8 − s = −8 ⓑ w + 25 − w + 5 = 0 w + 25 − w + 5 = 0

3 2 x − 3 − 20 = 7 3 2 x − 3 − 20 = 7

2 5 x + 1 − 8 = 0 2 5 x + 1 − 8 = 0

2 8 r + 1 − 8 = 2 2 8 r + 1 − 8 = 2

3 7 y + 1 − 10 = 8 3 7 y + 1 − 10 = 8

3 u − 2 = 5 u + 1 3 u − 2 = 5 u + 1

4 v + 3 = v − 6 4 v + 3 = v − 6

8 + 2 r = 3 r + 10 8 + 2 r = 3 r + 10

12 c + 6 = 10 − 4 c 12 c + 6 = 10 − 4 c

ⓐ a + 2 = a + 4 a + 2 = a + 4 ⓑ b − 2 + 1 = 3 b + 2 b − 2 + 1 = 3 b + 2

ⓐ r + 6 = r + 8 r + 6 = r + 8 ⓑ s − 3 + 2 = s + 4 s − 3 + 2 = s + 4

ⓐ u + 1 = u + 4 u + 1 = u + 4 ⓑ n − 5 + 4 = 3 n + 7 n − 5 + 4 = 3 n + 7

ⓐ x + 10 = x + 2 x + 10 = x + 2 ⓑ y − 2 + 2 = 2 y + 4 y − 2 + 2 = 2 y + 4

2 y + 4 + 6 = 0 2 y + 4 + 6 = 0

8 u + 1 + 9 = 0 8 u + 1 + 9 = 0

a + 1 = a + 5 a + 1 = a + 5

d − 2 = d − 20 d − 2 = d − 20

6 s + 4 = 8 s − 28 6 s + 4 = 8 s − 28

9 p + 9 = 10 p − 6 9 p + 9 = 10 p − 6

In the following exercises, solve. Round approximations to one decimal place.

Landscaping Reed wants to have a square garden plot in his backyard. He has enough compost to cover an area of 75 square feet. Use the formula s = A s = A to find the length of each side of his garden. Round your answer to the nearest tenth of a foot.

Landscaping Vince wants to make a square patio in his yard. He has enough concrete to pave an area of 130 square feet. Use the formula s = A s = A to find the length of each side of his patio. Round your answer to the nearest tenth of a foot.

Gravity While putting up holiday decorations, Renee dropped a light bulb from the top of a 64 foot tall tree. Use the formula t = h 4 t = h 4 to find how many seconds it took for the light bulb to reach the ground.

Gravity An airplane dropped a flare from a height of 1024 feet above a lake. Use the formula t = h 4 t = h 4 to find how many seconds it took for the flare to reach the water.

Gravity A hang glider dropped his cell phone from a height of 350 feet. Use the formula t = h 4 t = h 4 to find how many seconds it took for the cell phone to reach the ground.

Gravity A construction worker dropped a hammer while building the Grand Canyon skywalk, 4000 feet above the Colorado River. Use the formula t = h 4 t = h 4 to find how many seconds it took for the hammer to reach the river.

Accident investigation The skid marks for a car involved in an accident measured 54 feet. Use the formula s = 24 d s = 24 d to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Accident investigation The skid marks for a car involved in an accident measured 216 feet. Use the formula s = 24 d s = 24 d to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Accident investigation An accident investigator measured the skid marks of one of the vehicles involved in an accident. The length of the skid marks was 175 feet. Use the formula s = 24 d s = 24 d to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Accident investigation An accident investigator measured the skid marks of one of the vehicles involved in an accident. The length of the skid marks was 117 feet. Use the formula s = 24 d s = 24 d to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Writing Exercises

Explain why an equation of the form x + 1 = 0 x + 1 = 0 has no solution.

• ⓐ Solve the equation r + 4 − r + 2 = 0 r + 4 − r + 2 = 0 .
• ⓑ Explain why one of the “solutions” that was found was not actually a solution to the equation.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/elementary-algebra-2e/pages/1-introduction

• Authors: Lynn Marecek, MaryAnne Anthony-Smith, Andrea Honeycutt Mathis
• Publisher/website: OpenStax
• Book title: Elementary Algebra 2e
• Publication date: Apr 22, 2020
• Location: Houston, Texas
• Book URL: https://openstax.org/books/elementary-algebra-2e/pages/1-introduction
• Section URL: https://openstax.org/books/elementary-algebra-2e/pages/9-6-solve-equations-with-square-roots

© Jul 24, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

Squares and Square Roots in Algebra

You might like to read our Introduction to Squares and Square Roots first.

To square a number, just multiply it by itself.

Example: What is 3 squared?

"Squared" is often written as a little 2 like this:

Square Root

A square root goes the other direction:

3 squared is 9, so a square root of 9 is 3

It is like asking:

What can I multiply by itself to get this?

Here is the definition:

A square root of x is a number r whose square is x :

The Square Root Symbol

This is the special symbol that means "square root", it is like a tick, and actually started hundreds of years ago as a dot with a flick upwards. It is called the radical , and always makes mathematics look important!

We can use it like this:

Example: What is √36 ?

Answer: 6 × 6 = 36, so √36 = 6

Negative Numbers

We can also square negative numbers.

Example: What is minus 5 squared ?

... but ... what does "minus 5 squared" mean? 

• square the 5, then do the minus?
• or square (−5)?

It isn't clear! And we get different answers:

• square the 5, then do the minus: −(5×5) = −25
• square (−5): (−5)×(−5) = +25

So let's make it clear by using "( )".

Example Corrected: What is (−5) 2 ?

(−5) × (−5) = 25

(because a negative times a negative gives a positive )

That was interesting!

When we square a negative number we get a positive result.

Exactly the same as when we square a positive number:

Now remember our definition of a square root?

And we just found that:

(+5) 2 = 25 (−5) 2 = 25

There are two numbers whose square makes 25

So both +5 and −5 are square roots of 25

Two Square Roots

There can be a positive and negative square root!

This is important to remember.

Example: Solve w 2 = a

w = √a   and   w = −√a

Principal Square Root

So if there are really two square roots, why do people say √25 = 5 ?

Because √ means the principal square root ... the one that isn't negative!

There are two square roots, but the symbol √ means just the principal square root .

The square roots of 36 are 6 and −6

But √36 = 6 (not −6)

The Principal Square Root is sometimes called the Positive Square Root (but it can be zero).

Plus-Minus Sign

In a nutshell, why is this important.

Why is this "plus or minus" important? Because we don't want to miss a solution!

Example: Solve x 2 − 9 = 0

The "±" tells us to include the "−3" answer also.

Example: Solve for x in (x − 3) 2 = 16

Check: (7−3) 2 = 4 2 = 16 Check: (−1−3) 2 = (−4) 2 = 16

Square Root of xy

When two numbers are multiplied within a square root, we can split it into a multiplication of two square roots like this:

√ xy = √ x √ y

but only when x and y are both greater than or equal to 0

Example: What is √(100×4) ?

And √ x √ y = √ xy :

Example: What is √8√2 ?

Example: what is √(−8 × −2) .

We seem to have fallen into some trap here!

We can use Imaginary Numbers , but that leads to a wrong answer of −4

Oh that's right ...

The rule only works when x and y are both greater than or equal to 0

So we can't use that rule here.

Instead just do it this way:

√(−8 × −2) = √16 = +4

Why does √ xy = √ x √ y ?

We can use the fact that squaring a square root gives us the original value back again:

(√ a ) 2 = a

Assuming a is not negative!

An Exponent of a Half

A square root can also be written as a fractional exponent of one-half:

√ x = x ½ but only for x greater than or equal to 0

How About the Square Root of Negatives?

The result is an Imaginary Number ... read that page to learn more.

Reset password New user? Sign up

Existing user? Log in

Square Roots

Already have an account? Log in here.

• Ashish Menon
• Ameya Daigavane
• Kai Hsien Boo
• Prince Loomba
• E Tyson Ewing III
• Sandeep Bhardwaj
• Hopfhen Shane
• Hamza Waseem
• Bakul Choudhary
• Department 8
• John Taylor
• Mahindra Jain
• Hem Shailabh Sahu
• Kai Dettman
• Kaushik Surai
• Rindell Mabunga
• Johnny Gérard
• Skanda Prasad
• Andrew Hayes

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 \(b-a\).

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

Problem Loading...

Note Loading...

Set Loading...

• Solutions Integral Calculator Derivative Calculator Algebra Calculator Matrix Calculator More...
• Graphing Line Graph Exponential Graph Quadratic Graph Sine Graph More...
• Calculators BMI Calculator Compound Interest Calculator Percentage Calculator Acceleration Calculator More...
• Geometry Pythagorean Theorem Calculator Circle Area Calculator Isosceles Triangle Calculator Triangles Calculator More...
• Tools Notebook Groups Cheat Sheets Worksheets Study Guides Practice Verify Solution

• Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Number Line Expanded Form Mean, Median & Mode
• Algebra Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Interval Notation Pi (Product) Notation Induction Prove That Logical Sets Word Problems
• Pre Calculus Equations Inequalities Scientific Calculator Scientific Notation Arithmetics Complex Numbers Coterminal Angle Polar/Cartesian Simultaneous Equations System of Inequalities Polynomials Rationales Functions Arithmetic & Comp. Coordinate Geometry Plane Geometry Solid Geometry Conic Sections Trigonometry
• Calculus Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Fourier Transform
• Functions Line Equations Functions Arithmetic & Comp. Conic Sections Transformation
• Linear Algebra Matrices Vectors
• Trigonometry Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify
• Statistics Mean Geometric Mean Quadratic Mean Average Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge Standard Normal Distribution
• Physics Mechanics
• Chemistry Chemical Reactions Chemical Properties
• Finance Simple Interest Compound Interest Present Value Future Value
• Economics Point of Diminishing Return
• Conversions Roman Numerals Radical to Exponent Exponent to Radical To Fraction To Decimal To Mixed Number To Improper Fraction Radians to Degrees Degrees to Radians Hexadecimal Scientific Notation Distance Weight Time Volume
• Pre Algebra
• Two-step without parentheses
• Two-step with parentheses
• Three/four steps without parentheses
• Three/four steps with parentheses
• Multi-step without parentheses
• Multi-step with parentheses
• Prime Factorization
• Negative Factors
• Positive Factors
• Odd Factors
• Even Factors
• Biggest Factor
• Equivalent Fractions
• Add, Subtract
• Add, Subtract Like Denominators
• Add, Subtract Unlike Denominators
• Multiply with Whole Number
• Divide with Whole Number
• Mixed Numbers
• Complex Fractions
• Improper Fractions
• Negative Reciprocal
• Long Addition
• Long Subtraction
• Long Multiplication
• Long Division
• Long Division with Remainder
• Add/Subtract
• Multiplication
• Decimal to Fraction
• Fraction to Decimal
• Negative Exponent
• Square Root
• Ratios & Proportions

Number Line

• Expanded Form
• Pre Calculus
• Linear Algebra
• Trigonometry
• Conversions

square-root-calculator

• Practice Makes Perfect Learning math takes practice, lots of practice. Just like running, it takes practice and dedication. If you want...

We want your feedback

Please add a message.

Message received. Thanks for the feedback.