# 11.7: Equations with Radicals on Both Sides

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

**Practice**Equations with Radicals on Both Sides

What if you had a radical equation like \begin{align*}\sqrt{x + 1} + 2 = \sqrt{2x - 5}\end{align*} with a radical sign on both sides? How could you find the solutions to this equation? After completing this Concept, you'll be able to solve radical equations like this one.

### Watch This

CK-12 Foundation: Special Cases with Radical Equations

### Guidance

Often equations have more than one radical expression. The strategy in this case is to start by isolating the most complicated radical expression and raise the equation to the appropriate power. We then repeat the process until all radical signs are eliminated.

#### Example A

*Find the real roots of the equation* \begin{align*}\sqrt{2x+1}-\sqrt{x-3}=2\end{align*}.

**Solution**

\begin{align*}\text{Isolate one of the radical expressions:} && \sqrt{2x+1}& =2+\sqrt{x-3}\\ \text{Square both sides:} && \left(\sqrt{2x+1}\right)^2& =\left(2+\sqrt{x-3}\right)^2\\ \text{Eliminate parentheses:} && 2x+1& =4+4\sqrt{x-3}+x-3\\ \text{Simplify:} && x& =4 \sqrt{x-3}\\ \text{Square both sides of the equation:} && x^2& =\left(4 \sqrt{x-3} \right)^2\\ \text{Eliminate parentheses:} && x^2& =16(x-3)\\ \text{Simplify:} && x^2& =16x-48\\ \text{Move all terms to one side of the equation:} && x^2-16x+48& =0\\ \text{Factor:} && (x-12)(x-4)& =0\\ \text{Solve:} && x& =12 \ \text{or} \ x=4\end{align*}

**Check:** \begin{align*}\sqrt{2(12)+1}-\sqrt{12-3}=\sqrt{25}-\sqrt{9}=5-3=2\end{align*}. The solution checks out.

\begin{align*}\sqrt{2(4)+1}-\sqrt{4-3}=\sqrt{9}-\sqrt{1}=3-1=2\end{align*} The solution checks out.

The equation has two solutions: \begin{align*}x=12\end{align*} and \begin{align*}x=4\end{align*}.

**Identify Extraneous Solutions to Radical Equations**

We saw in Example 3 that some of the solutions that we find by solving radical equations do not check out when we substitute (or “plug in”) those solutions back into the original radical equation. These are called **extraneous solutions.** It is very important to check the answers we obtain by plugging them back into the original equation, so we can tell which of them are real solutions.

#### Example B

*Find the real solutions of the equation* \begin{align*}\sqrt{x-3}-\sqrt{x}=1\end{align*}.

**Solution**

\begin{align*}\text{Isolate one of the radical expressions:} && \sqrt{x-3}&=\sqrt{x}+1\\ \text{Square both sides:} && \left(\sqrt{x-3}\right)^2& =\left(\sqrt{x}+1\right)^2\\ \text{Remove parenthesis:} && x-3& =\left(\sqrt{x}\right)^2+2\sqrt{x}+1\\ \text{Simplify:} && x-3& =x+2\sqrt{x}+1\\ \text{Now isolate the remaining radical:} && -4& =2\sqrt{x}\\ \text{Divide all terms by 2:} && -2& =\sqrt{x}\\ \text{Square both sides:} && x& =4\end{align*}

**Check:** \begin{align*}\sqrt{4-3}-\sqrt{4}=\sqrt{1}-2=1-2=-1\end{align*} The solution does not check out.

** The equation has no real solutions.** \begin{align*}x=4\end{align*} is an extraneous solution.

**Applications using Special Case of Radical Equations**

Radical equations often appear in problems involving areas and volumes of objects.

#### Example C

*Anita’s square vegetable garden is 21 square feet larger than Fred’s square vegetable garden. Anita and Fred decide to pool their money together and buy the same kind of fencing for their gardens. If they need 84 feet of fencing, what is the size of each garden?*

**Solution**

**Make a sketch:**

**Define variables:** Let Fred’s area be \begin{align*}x\end{align*}; then Anita’s area is \begin{align*}x+21\end{align*}.

**Find an equation:**

Side length of Fred’s garden is \begin{align*}\sqrt{x}\end{align*}

Side length of Anita’s garden is \begin{align*}\sqrt{x+21}\end{align*}

The amount of fencing is equal to the combined perimeters of the two squares:

\begin{align*}4\sqrt{x}+4\sqrt{x+21}=84\end{align*}

**Solve the equation:**

\begin{align*}\text{Divide all terms by 4:} && \sqrt{x}+\sqrt{x+21}& =21\\ \text{Isolate one of the radical expressions:} && \sqrt{x+21}& =21-\sqrt{x}\\ \text{Square both sides:} && \left(\sqrt{x+21}\right)^2& =\left(21-\sqrt{x}\right)^2\\ \text{Eliminate parentheses:} && x+21& =441-42\sqrt{x}+x\\ \text{Isolate the radical expression:} && 42\sqrt{x}& =420\\ \text{Divide both sides by 42:} && \sqrt{x}& =10\\ \text{Square both sides:} && x& =100 \ ft^2\end{align*}

**Check:** \begin{align*}4\sqrt{100}+4\sqrt{100+21}=40+44=84\end{align*}. **The solution checks out.**

Fred’s garden is \begin{align*}10 \ ft \times 10 \ ft = 100 \ ft^2\end{align*} and Anita’s garden is \begin{align*}11 \ ft \times 11 \ ft = 121 \ ft^2\end{align*}.

Watch this video for help with the Examples above.

CK-12 Foundation: Special Cases with Radical Equations

### Vocabulary

- The symbol for a square root is \begin{align*}\sqrt{\;\;}\end{align*}. This symbol is also called the
**radical sign.**

### Guided Practice

*Find the real solutions of the equation* \begin{align*}\sqrt{9-x}=3+\sqrt{2x}\end{align*}.

**Solution**

\begin{align*}\text{Isolate one of the radical expressions:} && \sqrt{9-x}&=3+\sqrt{2x}\\ \text{Square both sides:} && (\sqrt{9-x})^2&=(3+\sqrt{2x})^2\\ \text{Remove parenthesis:} && 9-x&=9+6\sqrt{2x}+2x\\ \text{Simplify:} && x& =-2\sqrt{2x}\\ \text{Square both sides:} && x^2& =4(2x)\\ \text{Simplify:} && x^2& =8x\\ \text{Set one side equal to zero:} && x^2-8x& =0\\ \text{Factor:} && x(x-8)& =0\\ \text{Use the zero product principle:} && x& =0 \text{ or } x=8\end{align*}

**Check:** First check \begin{align*}x=0\end{align*}:

\begin{align*}\text{Start with the original equation:} && \sqrt{9-x}&=3+\sqrt{2x}\\ \text{Substitute in the x-value:} && \sqrt{9-8}&=3+\sqrt{2(8)}\\ \text{Simplify:} && \sqrt{1}&=3+\sqrt{16)}\\ \text{Take the square root:} && 1&=3+4=7\\ \end{align*}

The solution does not check out.

Then check \begin{align*}x=8\end{align*}:

\begin{align*}\text{Start with the original equation:} && \sqrt{9-x}&=3+\sqrt{2x}\\ \text{Substitute in the x-value:} && \sqrt{9-0}&=3+\sqrt{2(0)}\\ \text{Simplify:} && \sqrt{9}&=3+\sqrt{0)}\\ \text{Take the square root:} && 3&=3+0=3\\ \end{align*}

The solution checks out.

** The equation has one real solution.** \begin{align*}x=8\end{align*} is an extraneous solution.

### Practice

Find the solution to each of the following radical equations. Identify extraneous solutions.

- \begin{align*}\sqrt{x}=\sqrt{x-9}+1\end{align*}
- \begin{align*}\sqrt{x}+2=\sqrt{3x-2}\end{align*}
- \begin{align*}5 \sqrt{x}=\sqrt{x+12}+6\end{align*}
- \begin{align*}\sqrt{10-5x}+\sqrt{1-x}=7\end{align*}
- \begin{align*}\sqrt{2x-2}-2\sqrt{x}+2=0\end{align*}
- \begin{align*}\sqrt{2x+5}-3\sqrt{2x-3}=\sqrt{2-x}\end{align*}
- \begin{align*}3\sqrt{x}-9=\sqrt{2x-14}\end{align*}
- \begin{align*}\sqrt{x+7}=\sqrt{x+4}+1\end{align*}
- \begin{align*}\sqrt{4x}=\sqrt{3-2x}-1\end{align*}
- \begin{align*}\sqrt{2x+5}+\sqrt{3-x}=10\end{align*}

Extraneous Solution

An extraneous solution is a solution of a simplified version of an original equation that, when checked in the original equation, is not actually a solution.Radical Expression

A radical expression is an expression with numbers, operations and radicals in it.### Image Attributions

Here you'll learn how to find the real roots and identify the extraneous solutions of radical equations that have radicals on both sides of the equal sign. You'll also solve real-world applications that involve such equations.