Suppose your teacher has instructed the members of your math class to work in pairs, and she has asked you to find the length of a line segment. You get \begin{align*}\sqrt{2x+6}\end{align*}

### Radical Equations

Solving radical equations is no different from solving linear or quadratic equations. Before you can begin to solve a radical equation, you must know how to cancel the radical. To do that, you must know its **inverse**.

Original Operation |
Inverse Operation |
---|---|

Cube Root |
Cubing (to the third power) |

Square Root |
Squaring (to the second power) |

Fourth Root |
Fourth power |

“\begin{align*}n\end{align*} |
“\begin{align*}n\end{align*} |

To solve a radical equation, you apply the solving equation steps you learned in previous sections, including the inverse operations for roots.

#### Let's solve for the unknown value:

Solve \begin{align*}\sqrt{2x-1}=5\end{align*}

The first operation that must be removed is the square root. Square both sides.

\begin{align*}\left ( \sqrt{2x-1} \right )^2&=5^2\\ 2x-1&=25\\ 2x&=26\\ x&=13\end{align*}

Remember to check your answer by substituting it into the original problem to see if it makes sense.

#### Extraneous Solutions

Not every solution of a radical equation will check in the original problem. This is called an **extraneous solution**. This means you can find a solution using algebra, but it will not work when checked. This is because of the rule in a previous section:

\begin{align*}\sqrt[n]{x}\end{align*} is undefined when \begin{align*}n\end{align*} is an even whole number and \begin{align*}x<0\end{align*}, or, in words, even roots of negative numbers are undefined.

#### Now, let's solve the following equation:

\begin{align*}\sqrt{x-3}-\sqrt{x}=1\end{align*}

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

Check: \begin{align*}\sqrt{4-3} \stackrel{?}{=} \sqrt{4}+1\Rightarrow \sqrt{1} \stackrel{?}{=} 2+1 \Rightarrow 1\neq 3 \end{align*}. The solution does not check out. The equation has no real solutions. Therefore, \begin{align*}x=4\end{align*} is an extraneous solution.

#### Finally, let's use what we learned about extraneous solutions to find the volume of a sphere.

A sphere has a volume of \begin{align*}456 \ cm^3\end{align*}. If the radius of the sphere is increased by 2 cm, what is the new volume of the sphere?

First, define the variables. Let \begin{align*}R=\end{align*} the radius of the sphere.

Now, find an equation. The volume of a sphere is given by the formula: \begin{align*}V=\frac{4}{3}\pi r^3\end{align*}.

By substituting 456 for the volume variable, the equation becomes \begin{align*}456=\frac{4}{3} \pi r^3\end{align*}.

\begin{align*}\text{Multiply by} \ 3: \qquad \quad 1368 &= 4\pi r^3\\ \text{Divide by} \ 4\pi: \qquad 108.92&=r^3\\ \text{Take the cube root of each side:} \qquad \qquad \ r&=\sqrt[3]{108.92} \Rightarrow r = 4.776 \ cm\\ \text{The new radius is 2 centimeters more:} \qquad \qquad \ r &= 6.776 \ cm\\ \text{The new volume is}: \qquad \qquad V &= \frac{4}{3}\pi (6.776)^3 = 1302.5 \ cm^3\end{align*}

Check by substituting the values of the radii into the volume formula.

\begin{align*}V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi (4.776)^3=456 \ cm^3\end{align*}. The solution checks out.

### Examples

#### Example 1

Earlier, you and a class partner are asked to find the length of a line segment. You get \begin{align*}\sqrt{2x+6}\end{align*} units for the length, and your partner gets 10 units. You ask your teacher who is right, and she says both of you are right! Can you set up an equation and solve for \begin{align*}x\end{align*} in this case? How would you do it?

Because both answers are right, you can set them equal to each other and solve for \begin{align*}x\end{align*}.

\begin{align*}&\sqrt{2x+6}=10\\ &{\sqrt{2x+6}}^{2}={10}^{2}\\ &2x+6=100\\ &2x=94\\ &x=47\\\end{align*}

If you set \begin{align*}x=47\end{align*}, then both you and your partner have the same answer.

#### Example 2

Solve \begin{align*}\sqrt{x+15}=\sqrt{3x-3}\end{align*}.

Begin by canceling the square roots by squaring both sides.

\begin{align*}\left ( \sqrt{x+15} \right )^2&=\left ( \sqrt{3x-3} \right )^2\\ x+15&=3x-3\\ \text{Isolate the} \ x-\text{variable}: \qquad \qquad 18 & = 2x\\ x&=9\end{align*}

Check the solution: \begin{align*}\sqrt{9+15}=\sqrt{3(9)-3} \rightarrow \sqrt{24}=\sqrt{24}\end{align*}. The solution checks out.

### Review

In 1-16, find the solution to each of the following radical equations. Identify extraneous solutions.

- \begin{align*}\sqrt{x+2}-2=0\end{align*}
- \begin{align*}\sqrt{3x-1}=5\end{align*}
- \begin{align*}2\sqrt{4-3x}+3=0\end{align*}
- \begin{align*}\sqrt[3]{x-3}=1\end{align*}
- \begin{align*}\sqrt[4]{x^2-9}=2\end{align*}
- \begin{align*}\sqrt[3]{-2-5x}+3=0\end{align*}
- \begin{align*}\sqrt{x}=x-6\end{align*}
- \begin{align*}\sqrt{x^2-5x}-6=0\end{align*}
- \begin{align*}\sqrt{(x+1)(x-3)}=x\end{align*}
- \begin{align*}\sqrt{x+6}=x+4\end{align*}
- \begin{align*}\sqrt{x}=\sqrt{x-9}+1\end{align*}
- \begin{align*}\sqrt{3x+4}=-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*}
- The area of a triangle is \begin{align*}24 \ in^2\end{align*} and the height of the triangle is twice as long and the base. What are the base and the height of the triangle?
- The volume of a square pyramid is given by the formula \begin{align*}V=\frac{A(h)}{3}\end{align*}, where \begin{align*}A=\end{align*} area of the base and \begin{align*}h=\end{align*} height of the pyramid. The volume of a square pyramid is 1,600 cubic meters. If its height is 10 meters, find the area of its base.
- The volume of a cylinder is \begin{align*}245 \ cm^3\end{align*} and the height of the cylinder is one-third the diameter of the cylinder's base. The diameter of the cylinder is kept the same, but the height of the cylinder is increased by two centimeters. What is the volume of the new cylinder? \begin{align*}(\text{Volume} = \pi r^2 \cdot h)\end{align*}
- The height of a golf ball as it travels through the air is given by the equation \begin{align*}h=-16t^2+256\end{align*}. Find the time when the ball is at a height of 120 feet.

**Mixed Review**

- Joy sells two types of yarn: wool and synthetic. Wool is $12 per skein and synthetic is $9 per skein. If Joy sold 16 skeins of synthetic and collected a total of $432, how many skeins of wool did she sell?
- Solve \begin{align*}16 \ge |x-4|\end{align*}.
- Graph the solution: \begin{align*}\begin{cases}y \le 2x-4\\ y>-\frac{1}{4} x+6\end{cases}\end{align*}.
- You randomly point to a day in the month of February, 2011. What is the probability your finger lands on a Monday?
- Carbon-14 has a half life of 5,730 years. Your dog dug a bone from your yard. It had 93% of its carbon-14 remaining. How old is the bone?
- What is true about solutions to inconsistent systems?

### Review (Answers)

To see the Review answers, open this PDF file and look for section 11.5.