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*}
Guidance
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 Concepts, including the inverse operations for roots.
Example A
Solve \begin{align*}\sqrt{2x1}=5\end{align*}
Solution:
The first operation that must be removed is the square root. Square both sides.
\begin{align*}\left ( \sqrt{2x1} \right )^2&=5^2\\
2x1&=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 Concept:
\begin{align*}\sqrt[n]{x}\end{align*}
or, in words, even roots of negative numbers are undefined.
Example B
Solve \begin{align*}\sqrt{x3}\sqrt{x}=1\end{align*}
Solution:
\begin{align*}\text{Isolate one of the radical expressions}. \qquad \ \sqrt{x3}&=\sqrt{x}+1\\
\text{Square both sides}. \quad \left ( \sqrt{x3} \right )^2 & = \left ( \sqrt{x}+1 \right )^2\\
\text{Remove parentheses}. \qquad \quad \ x3&=\left ( \sqrt{x} \right )^2 + 2\sqrt{x}+1\\
\text{Simplify}. \qquad \quad \ x3&=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{43} \stackrel{?}{=} \sqrt{4}+1\Rightarrow \sqrt{1} \stackrel{?}{=} 2+1 \Rightarrow 1\neq 3 \end{align*}
Radical Equations in Real Life
Example C
A sphere has a volume of \begin{align*}456 \ cm^3\end{align*}
Solution:

Define variables. Let \begin{align*}R=\end{align*}
R= the radius of the sphere. 
Find an equation. The volume of a sphere is given by the formula: \begin{align*}V=\frac{4}{3}\pi r^3\end{align*}
V=43πr3 .
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*}
Guided Practice
Solve \begin{align*}\sqrt{x+15}=\sqrt{3x3}\end{align*}
Solution:
Begin by canceling the square roots by squaring both sides.
\begin{align*}\left ( \sqrt{x+15} \right )^2&=\left ( \sqrt{3x3} \right )^2\\
x+15&=3x3\\
\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*}
Practice
Sample explanations for some of the practice exercises below are available by viewing the following videos. Note that there is not always a match between the number of the practice exercise in the videos and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK12 Basic Algebra: Extraneous Solutions to Radical Equations (11:10)
CK12 Basic Algebra: Radical Equation Examples (5:16)
CK12 Basic Algebra: More Involved Radical Equation Example (11:54)
In 116, find the solution to each of the following radical equations. Identify extraneous solutions.

\begin{align*}\sqrt{x+2}2=0\end{align*}
x+2−−−−−√−2=0 
\begin{align*}\sqrt{3x1}=5\end{align*}
3x−1−−−−−√=5 
\begin{align*}2\sqrt{43x}+3=0\end{align*}
24−3x−−−−−√+3=0 
\begin{align*}\sqrt[3]{x3}=1\end{align*}
x−3−−−−−√3=1 
\begin{align*}\sqrt[4]{x^29}=2\end{align*}
x2−9−−−−−√4=2 
\begin{align*}\sqrt[3]{25x}+3=0\end{align*}
−2−5x−−−−−−−√3+3=0 
\begin{align*}\sqrt{x}=x6\end{align*}
x√=x−6 
\begin{align*}\sqrt{x^25x}6=0\end{align*}
x2−5x−−−−−−√−6=0 
\begin{align*}\sqrt{(x+1)(x3)}=x\end{align*}
(x+1)(x−3)−−−−−−−−−−−√=x 
\begin{align*}\sqrt{x+6}=x+4\end{align*}
x+6−−−−−√=x+4 
\begin{align*}\sqrt{x}=\sqrt{x9}+1\end{align*}
x√=x−9−−−−−√+1 
\begin{align*}\sqrt{3x+4}=6\end{align*}
3x+4−−−−−√=−6 
\begin{align*}\sqrt{105x}+\sqrt{1x}=7\end{align*}
10−5x−−−−−−√+1−x−−−−−√=7 
\begin{align*}\sqrt{2x2}2\sqrt{x}+2=0\end{align*}
2x−2−−−−−√−2x√+2=0 
\begin{align*}\sqrt{2x+5}3\sqrt{2x3}=\sqrt{2x}\end{align*}
2x+5−−−−−√−32x−3−−−−−√=2−x−−−−−√ 
\begin{align*}3\sqrt{x}9=\sqrt{2x14}\end{align*}
3x√−9=2x−14−−−−−−√  The area of a triangle is \begin{align*}24 \ in^2\end{align*}
24 in2 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*}
V=A(h)3 , where \begin{align*}A=\end{align*}A= area of the base and \begin{align*}h=\end{align*}h= 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*}
245 cm3 and the height of the cylinder is onethird 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*}(Volume=πr2⋅h)  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*}
h=−16t2+256 . 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 x4\end{align*}
16≥x−4 .  Graph the solution: \begin{align*}\begin{cases}y \le 2x4\\ 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?
 Carbon14 has a half life of 5,730 years. Your dog dug a bone from your yard. It had 93% of its carbon14 remaining. How old is the bone?
 What is true about solutions to inconsistent systems?