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9.4: Equations with Square Roots

Difficulty Level: At Grade Created by: CK-12

Let’s Think About It

License: CC BY-NC 3.0

Running from one base to the next was a speedy 90 feet for Omar. The job of mowing the field, however, probably wasn’t so fast. Just counting the infield, how many square feet of mowing does the groundskeeper do?

In this concept, you will learn how to solve equations using squares and square roots.


You may already know that squaring a number and taking the square root of a number are opposite operations. If you know one, you can find the other. When working with area and dimensions, the equations most often used are:

\begin{align*}A=s^2\end{align*}A=s2 when you know the sides and need to find the area.

\begin{align*}s= \sqrt{A}\end{align*}s=A when you know the area and want to find the length of the sides.

In these equations, both \begin{align*}s\end{align*}s and \begin{align*}A\end{align*}A are variables. A variable is simply an unknown quantity represented by a letter. Any letter or symbol can be used in a math sentence as a variable. For example, the equation \begin{align*}y=x^2\end{align*}y=x2 says that some number, \begin{align*}y\end{align*}y, is equal to another number, \begin{align*}x\end{align*}x, times itself. The equation \begin{align*}x=\sqrt{y}\end{align*}x=y is the opposite operation and says some number, \begin{align*}x\end{align*}x, is equal to the square root of another number, \begin{align*}y\end{align*}y.

Here is an example of how to use these equations to solve problems involving squares and square roots.

Solve for \begin{align*}y\end{align*}y:


First, you know that \begin{align*}5^2\end{align*}52 is the same as \begin{align*}5 \times 5\end{align*}5×5

Next, \begin{align*}5 \times 5=25\end{align*}5×5=25

Then, \begin{align*}y=25\end{align*}y=25

Your answer is 25.

Here’s another example:

Solve for \begin{align*}x\end{align*}x:

\begin{align*}x^2 = 36\end{align*}x2=36

First, you know that \begin{align*}x\end{align*}x is the unknown variable that you are looking for.

You may also know that in order to find \begin{align*}x\end{align*}x in any equation, you need to get it on either side of the equal sign, by ITSELF. This means it cannot have anything else attached to it by any other operation, including squares.

Next, in order to isolate \begin{align*}x\end{align*}x, you must perform the opposite operation. Ask yourself, “WHAT is attached to the \begin{align*}x\end{align*}x?” and “HOW is it attached?”

Since a square is attached to the \begin{align*}x\end{align*}x, the square root of \begin{align*}x\end{align*}x is the opposite operation.

Then, remember, that whatever you do to one side of the equation, you must also do to the other side.

Take the square root of both sides of the equation.

\begin{align*}\sqrt{x^2} = \sqrt{36}\end{align*}x2=36

Since you have been given an abstract problem to solve, be sure to include negative roots.

\begin{align*}x = \pm 6\end{align*}x=±6

The answer is \begin{align*}\pm 6\end{align*}±6

Guided Practice


\begin{align*}x^2 + 3 = 12\end{align*}x2+3=12

First, recognize that you are solving for \begin{align*}x\end{align*}x and determine the best way to isolate it. In this case there is more than one function attached, a square and a 3 by addition.

Next, determine which function to remove first from the \begin{align*}x\end{align*}x. Since you must do the same operation to everything on both sides of the equation, the 3 is the easiest to remove first. You do this by subtraction because it is attached by addition.

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


After subtracting, you have a new equation:

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


Then, all that is left is to take the square root of both sides.

\begin{align*}\begin{array}{rcl} \sqrt{x^2} &=& \sqrt{9} \\ x &=& \pm 3 \end{array}\end{align*}


The answer is \begin{align*}\pm 3\end{align*}±3


Example 1

Solve for \begin{align*}y\end{align*}y:

\begin{align*}\sqrt{y-1} = 8\end{align*}y1=8

First, recognize the two operations attached to the variable - a 1 by subtraction and a square root.

Next, determine which operation you can perform first to both sides as a step toward isolating \begin{align*}y\end{align*}y.

Perform the opposite of a square root by squaring both sides of the equation.

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


This gives you a new equation:

\begin{align*}y-1=64 \end{align*}


Then add to remove the 1 that is attached by subtraction.

\begin{align*}\begin{array}{rcl} y -1 +1 &=& 64 + 1 \\ y &=& 64 +1 \\ y &=& 65 \end{array}\end{align*}


The answer is 65.

Example 2

Solve for \begin{align*}x\end{align*}x:


First, take the square root of both sides.

\begin{align*}\sqrt{x^2} = \sqrt{49}\end{align*}x2=49

Then, \begin{align*}x= \pm 7\end{align*}x=±7

The answer is \begin{align*}\pm 7\end{align*}±7

Example 3

Solve for \begin{align*}p\end{align*}p:

\begin{align*}p^2 + 5=174\end{align*}p2+5=174

First, subtract 5 from both sides of the equation.

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

Next, rewrite the equation:

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

Then, take the square root of both sides.

\begin{align*}p=\pm 13\end{align*}

The answer is \begin{align*}p=\pm 13\end{align*}

Follow Up

License: CC BY-NC 3.0

Remember Omar wondering about how many square feet of grass needed to be mowed?

One side of the square baseball diamond infield measured 90 feet.

First, remember the formula for area of a square.


Next, substitute in what you know.

\begin{align*}A=(90 \ ft)^2\end{align*}

Then solve for \begin{align*}A\end{align*}.

\begin{align*}90 \ ft \times 90 \ ft = 1,800 \ sq \ ft\end{align*}

The answer is 1,800 square feet. Remember that area is measured in squares.

Video Review


Explore More

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 = 225\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*}

Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 9.4.  

Image Attributions

  1. [1]^ License: CC BY-NC 3.0
  2. [2]^ License: CC BY-NC 3.0


Difficulty Level:

At Grade


Date Created:

Dec 02, 2015

Last Modified:

Dec 02, 2015
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