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- Mathematics

## How to Solve Square Root Problems

Last Updated: November 3, 2022 References

## Understanding Squares and Square Roots

- 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 .

- 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 .

## 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.

## Quickly Estimating Imperfect Squares

- 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.

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## 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

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## Unit 10: Lesson 2

- Intro to square-root equations & extraneous solutions
- Square-root equations intro
- Intro to solving square-root equations

## Solving square-root equations

- Solving square-root equations: one solution
- Solving square-root equations: two solutions
- Solving square-root equations: no solution

## Introduction

Practice question 1: isolating the radical term.

- Your answer should be
- an integer, like 6 6 6 6
- a simplified proper fraction, like 3 / 5 3/5 3 / 5 3, slash, 5
- a simplified improper fraction, like 7 / 4 7/4 7 / 4 7, slash, 4
- a mixed number, like 1 3 / 4 1\ 3/4 1 3 / 4 1, space, 3, slash, 4
- an exact decimal, like 0.75 0.75 0 . 7 5 0, point, 75
- a multiple of pi, like 12 pi 12\ \text{pi} 1 2 pi 12, space, start text, p, i, end text or 2 / 3 pi 2/3\ \text{pi} 2 / 3 pi 2, slash, 3, space, start text, p, i, end text

## Practice question 2: Two possible solutions

- (Choice A) x = 0 x=0 x = 0 x, equals, 0 A x = 0 x=0 x = 0 x, equals, 0
- (Choice B) x = − 4 x=-4 x = − 4 x, equals, minus, 4 B x = − 4 x=-4 x = − 4 x, equals, minus, 4
- (Choice C) x = 4 x=4 x = 4 x, equals, 4 C x = 4 x=4 x = 4 x, equals, 4
- (Choice D) x = − 6 x=-6 x = − 6 x, equals, minus, 6 D x = − 6 x=-6 x = − 6 x, equals, minus, 6
- (Choice E) x = 1 x=1 x = 1 x, equals, 1 E x = 1 x=1 x = 1 x, equals, 1

## Practice question 3

- (Choice B) x = − 5 x=-5 x = − 5 x, equals, minus, 5 B x = − 5 x=-5 x = − 5 x, equals, minus, 5
- (Choice C) x = 1 x=1 x = 1 x, equals, 1 C x = 1 x=1 x = 1 x, equals, 1
- (Choice D) x = − 5 2 x=-\dfrac{5}{2} x = − 2 5 x, equals, minus, start fraction, 5, divided by, 2, end fraction D x = − 5 2 x=-\dfrac{5}{2} x = − 2 5 x, equals, minus, start fraction, 5, divided by, 2, end fraction
- (Choice E) None of the above E None of the above

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## Square Root Calculator

## Calculator Use

## Square Roots, odd and even:

## Perfect Square Calculator

## Example Square Roots:

- The 2nd root of 81, or 81 radical 2, or the square root of 81 is written as $$ \sqrt[2]{81} = \sqrt[]{81} = \pm 9 $$.
- The 2nd root of 25, or 25 radical 2, or the square root of 25 is written as $$ \sqrt[2]{25} = \sqrt[]{25} = \pm 5 $$.
- The 2nd root of 100, or 100 radical 2, or the square root of 100 is written as $$ \sqrt[2]{100} = \sqrt[]{100} = \pm 10 $$.
- The 2nd root of 10, or 10 radical 2, or the square root of 10 is written as $$ \sqrt[2]{10} = \sqrt[]{10} = \pm 3.162278 $$.

To calculate fractional exponents use our calculator for Fractional Exponents .

[1] Weisstein, Eric W. "Square Root." From MathWorld -- A Wolfram Web Resource. Square Root

## Square Roots Calculator

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## 9.6: Solve Equations with Square Roots

## Learning Objectives

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

Before you get started, take this readiness quiz.

- Simplify: ⓐ \(\sqrt{9}\) ⓑ \(9^2\). If you missed this problem, review Example 9.1.1 and Exercise 1.3.22 .
- Solve: 5(x+1)−4=3(2x−7). If you missed this problem, review Exercise 2.4.16 .
- Solve: \(n^2−6n+8=0\). If you missed this problem, review Exercise 7.6.13 .

## Solve Radical Equations

## Definition: RADICAL EQUATION

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

## Example \(\PageIndex{1}\)

For the equation \(\sqrt{x+2}=x\):

## Example \(\PageIndex{2}\)

For the equation \(\sqrt{x+6}=x\):

## Example \(\PageIndex{3}\)

For the equation \(\sqrt{−x+2}=x\):

For\(a \ge 0\), \((\sqrt{a})^2=a\)

How to Solve Radical Equations

## Example \(\PageIndex{4}\)

## Example \(\PageIndex{5}\)

## Example \(\PageIndex{6}\)

## Definition: SOLVE A RADICAL EQUATION.

- Isolate the radical on one side of the equation.
- Square both sides of the equation.
- Solve the new equation.
- Check the answer.

## Example \(\PageIndex{7}\)

## Example \(\PageIndex{8}\)

## Example \(\PageIndex{9}\)

## Example \(\PageIndex{10}\)

## Example \(\PageIndex{11}\)

## Example \(\PageIndex{12}\)

## Example \(\PageIndex{13}\)

## Example \(\PageIndex{14}\)

## Example \(\PageIndex{15}\)

## Definition: BINOMIAL SQUARES

\[\begin{array}{cc} {(a+b)^2=a^2+2ab+b^2}&{(a−b)^2=a^2−2ab+b^2}\\ \nonumber \end{array}\]

## Example \(\PageIndex{16}\)

## Example \(\PageIndex{17}\)

## Example \(\PageIndex{18}\)

## Example \(\PageIndex{19}\)

## Example \(\PageIndex{20}\)

## Example \(\PageIndex{21}\)

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

## Example \(\PageIndex{22}\)

## Example \(\PageIndex{23}\)

Solve: \(\sqrt{24a+2}−16=16\).

## Example \(\PageIndex{24}\)

Solve: \(\sqrt{36b+3}−25=50\).

## Example \(\PageIndex{25}\)

Solve: \(\sqrt{4z−3}=\sqrt{3z+2}\).

## Example \(\PageIndex{26}\)

Solve: \(\sqrt{2x−5}=\sqrt{5x+3}\).

## Example \(\PageIndex{27}\)

Solve: \(\sqrt{7y+1}=\sqrt{2y−5}\).

## Example \(\PageIndex{28}\)

Solve: \(\sqrt{m}+1=\sqrt{m+9}\).

## Example \(\PageIndex{29}\)

Solve: \(\sqrt{x}+3=\sqrt{x+5}\).

## Example \(\PageIndex{30}\)

Solve: \(\sqrt{m}+5=\sqrt{m+16}\).

## Example \(\PageIndex{31}\)

Solve: \(\sqrt{q−2}+3=\sqrt{4q+1}\).

## Example \(\PageIndex{32}\)

Solve: \(\sqrt{y−3}+2=\sqrt{4y+2}\).

## Example \(\PageIndex{33}\)

Solve: \(\sqrt{n−4}+5=\sqrt{3n+3}\).

## Use Square Roots in Applications

## Definition: SOLVE APPLICATIONS WITH FORMULAS.

- 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.
- Identify what we are looking for.
- Name what we are looking for by choosing a variable to represent it.
- Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
- Solve the equation using good algebra techniques.
- Check the answer in the problem and make sure it makes sense.
- Answer the question with a complete sentence.

We can use the formula \(s=\sqrt{A}\) to find the length of a side of a square for a given area.

## Definition: AREA OF A SQUARE

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

## Example \(\PageIndex{34}\)

## Example \(\PageIndex{35}\)

## Example \(\PageIndex{36}\)

Another application of square roots has to do with gravity.

## Definition: FALLING OBJECTS

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

## Example \(\PageIndex{37}\)

## Exercise \(\PageIndex{38}\)

## Example \(\PageIndex{39}\)

## Definition: SKID MARKS AND SPEED OF A CAR

## Example \(\PageIndex{40}\)

## Example \(\PageIndex{41}\)

## Example \(\PageIndex{42}\)

## Key Concepts

- On Earth, if an object is dropped from a height of hh feet, the time in seconds it will take to reach the ground is found by using the formula\(t=\frac{\sqrt{h}}{4}\).
- 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 \(s=\sqrt{24d}\).

## IMAGES

## VIDEO

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