# 9.4: Equations with Square Roots

**At Grade**Created by: CK-12

**Practice**Equations with Square Roots

Have you ever thought about the square footage of the infield in baseball?

The infield is the area of the baseball field that has the base lines as boundaries. It is the area contained within the base lines. The area outside of the base lines is called the outfield. We know that the distance from one base to another is \begin{align*}90 \ ft\end{align*}

\begin{align*}y = x^2\end{align*}

What is \begin{align*}x\end{align*}

What is \begin{align*}y\end{align*}

**Solving this equation requires that you understand how to solve equations by using square roots. Pay attention and we will return to this equation at the end of the Concept.**

### Guidance

If we know that the square of a number is equal to a product, then we can write this equation.

\begin{align*}y = x^2\end{align*}

This is an equation that expresses that if we multiply the value of \begin{align*}x\end{align*}

**We can say that \begin{align*}x\end{align*} x is the square root of \begin{align*}y\end{align*}y.**

**We can use this information to solve equations involving square roots.**

\begin{align*}x^2=81\end{align*}

**To solve this equation, we want to figure out the value of \begin{align*}x\end{align*} x. To do this, we can take the square root of both sides of the equation.**

**Why would we do this?**

Think about it. The \begin{align*}x\end{align*}

\begin{align*}\sqrt{x^2} = \sqrt{81}\end{align*}

Next, we can cancel the square and the square root. They are inverses of each other and they cancel each other out.

\begin{align*}\bcancel{\sqrt{x^2}} = x\end{align*}

Now we find the square root of 81. 81 is a perfect square, so the square root of 81 is 9.

\begin{align*}x=9\end{align*}

**This is our answer.**

**Sometimes, you will have a problem that is a little more complicated. Take a look.**

\begin{align*}x^2+3=12\end{align*}

We want to find the value of \begin{align*}x\end{align*}

Let’s start by subtracting three from both sides.

\begin{align*}x^2+3-3&=12-3\\
x^2&=9\end{align*}

Now we want to get \begin{align*}x\end{align*}

\begin{align*}\bcancel{\sqrt{x^2}} &= \sqrt{9}\\
x&=3\end{align*}

**This is our answer.**

**Sometimes, the equation with have a square root in it and we have to work with that.**

\begin{align*}\sqrt{x-1} = 8\end{align*}

Wow! Here we have a variable and a number in a radical. Let’s get rid of the radical first.

To get cancel out the square root of a number, we use the inverse operation. We square both sides of the equation.

\begin{align*}\left ( \sqrt{x-1} \right )^2 = 8^2\end{align*}

The square and the square root cancel each other out.

\begin{align*}(\bcancel{\sqrt{x-1})^2} &= 8^2\\
x-1 &= 64\end{align*}

Now we can solve for \begin{align*}x\end{align*}

\begin{align*}x-1+1&=64+1\\
x&=65\end{align*}

**This is our answer.**

Now it's time for you to try a few on your own.

#### Example A

\begin{align*}x^2=49\end{align*}

**Solution:\begin{align*}x = 7\end{align*} x=7**

#### Example B

\begin{align*}x^2=64\end{align*}

**Solution:\begin{align*}x = 8\end{align*} x=8**

#### Example C

\begin{align*}x^2 + 2=38\end{align*}

**Solution:\begin{align*}x = 6\end{align*} x=6**

Here is the original problem once again.

The infield is the area of the baseball field that has the base lines as boundaries. It is the area contained within the base lines. The area outside of the base lines is called the outfield. We know that the distance from one base to another is \begin{align*}90 \ ft\end{align*}

\begin{align*}y = x^2\end{align*}

What is \begin{align*}x\end{align*}

What is \begin{align*}y\end{align*}

Let's begin by looking at the given information. In the original problem, it is presented that the distance from first base to second base times the distance from second base to third base would give the area of the infield. The distance from first to second base is the same as the distance from second to third. This is our \begin{align*}x\end{align*}

\begin{align*}x = 90 \ ft\end{align*}

Now we can substitute this value for \begin{align*}x\end{align*}

\begin{align*}y = 90^2\end{align*}

\begin{align*}y = 8100 \ sq. ft.\end{align*}

**This is our answer.**

### Vocabulary

Here are the vocabulary words that are found in this Concept.

- Square root
- a number that when multiplied by itself equals the square of the number.

- Perfect Square
- square roots that are whole numbers.

- Radical
- the symbol that lets us know that we are looking for a square root.

- Radical Expression
- an expression with numbers, operations and radicals in it.

### Guided Practice

Here is one for you to try on your own.

Evaluate.

\begin{align*}x^2+5=174\end{align*}

**Answer**

To start, let's subtract five from both sides.

\begin{align*}x^2+5-5=174-5\end{align*}

\begin{align*}x^2 = 169\end{align*}

Now we take the square root of both sides.

\begin{align*}\sqrt{x^2} = \sqrt{169}\end{align*}

\begin{align*}x = 13\end{align*}

**This is our answer.**

### Video Review

Here is a video for review.

This is a James Sousa video on solving equations with one radical.

### Practice

Directions: Solve each equation.

1. \begin{align*}x^2=9\end{align*}

2. \begin{align*}x^2=49\end{align*}

3. \begin{align*}x^2=100\end{align*}

4. \begin{align*}x^2=64\end{align*}

5. \begin{align*}x^2=90\end{align*}

6. \begin{align*}x^2=256\end{align*}

7. \begin{align*}x^2+3=12\end{align*}

8. \begin{align*}x^2-5=20\end{align*}

9. \begin{align*}x^2+3=39\end{align*}

10. \begin{align*}x^2-4=60\end{align*}

11. \begin{align*}x^2+11=92\end{align*}

12. \begin{align*}\sqrt{x+1} = 10\end{align*}

13. \begin{align*}x^2+5=41\end{align*}

14. \begin{align*}x^3=8\end{align*}

15. \begin{align*}x^3+4=31\end{align*}

### Notes/Highlights Having trouble? Report an issue.

Color | Highlighted Text | Notes | |
---|---|---|---|

Show More |

### Image Attributions

Here you'll learn to solve equations using square roots.