<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 9.4: Equations with Square Roots

Difficulty Level: At Grade Created by: CK-12
Estimated3 minsto complete
%
Progress
Practice Equations with Square Roots

MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated3 minsto complete
%
Estimated3 minsto complete
%
MEMORY METER
This indicates how strong in your memory this concept is

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 90 ft\begin{align*}90 \ ft\end{align*}. The shape of the infield is a square. It makes sense to say that the distance from first base to second base times the distance from second base to third base will provide us with the measurement for the area of the infield. We can write the following equation.

\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*} by itself, we will end up with the value of \begin{align*}y\end{align*}.

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

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*}. 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*} is being squared. We want to get the variable alone. To do this, we need to perform the inverse operation. The opposite of squaring a number is finding the square root of the number. Therefore, if we take the square root of both sides of the equation, then we will get \begin{align*}x\end{align*} alone.

\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*}

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*}. First, notice that we have a two step equation. One of the operations is multiplication with the square and the other is addition.

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*} alone. To do this, we take the square root of both sides of the equation.

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

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*} quite easily. Begin by adding 1 to both sides of the equation.

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

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*}

#### Example B

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

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

#### Example C

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

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

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*}. The shape of the infield is a square. It makes sense to say that the distance from first base to second base times the distance from second base to third base will provide us with the measurement for the area of the infield. We can write the following equation.

\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*} measurement.

\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*}

### 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.
the symbol that lets us know that we are looking for a square root.
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*}

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*}

### Video Review

Here is a video for review.

### 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 Hide Details
Description
Difficulty Level:
Authors:
Tags:
Subjects: