

A radical equation is any equation that contains one or more radicals with a variable in the radicand. Following are some examples of radical equations, all of which will be solved in this section:

 $$\sqrt { 2 x - 1 } = 3$$ $$\sqrt [ 3 ] { 4 x ^ { 2 } + 7 } - 2 = 0$$ $$\sqrt { x + 2 } - \sqrt { x } = 1$$

The squaring property of equality states that if given real numbers $$a$$ and $$b$$ that are equal, the equality is retained if both numbers are squared. For example.

Given $$- 3 = - 3$$, then squaring both quantities is also a true statement: $$( - 3 ) ^ { 2 } = ( - 3 ) ^ { 2 }$$ because  $$9 = 9\:\:\color{Cerulean}{✓}$$

The converse, on the other hand, is not necessarily true,

Given $$9 = 9$$ which could be written $$( - 3 ) ^ { 2 }= ( 3 ) ^ { 2 }$$ does not produce an equality if the squaring operation is removed: $$- 3 \neq 3\:\:\color{red}{✗}$$

This is important because we will use this property to solve radical equations. Because the converse of the squaring property of equality is not necessarily true, solutions to the squared equation may not be solutions to the original. Hence squaring both sides of an equation introduces the possibility of extraneous solutions, which are solutions that do not solve the original equation. For example, to find the solution to $$\sqrt { x } = - 5$$, the technique used is to square both sides of the equal sign,$${\color{Cerulean}{(}} \sqrt { x } {\color{Cerulean}{) ^ { 2 }}} = {\color{Cerulean}{ (}}5 {\color{Cerulean}{ ) ^ { 2 }}}$$ which produces the solution $$x = 25$$. However, checking this solution produces $$\sqrt { 25 } = 5$$ which contradicts the original problem statement, $$\sqrt { x } = - 5$$.

For this reason, answers that result from squaring both sides of an equation must ALWAYS be checked.

How to: Solve a Radical Equation.

1. Isolate a radical. Put ONE radical on one side of the equal sign and put everything else on the other side.
2. Eliminate the radical. Raise both sides of the equal sign to the power that matches the index on the radical. This means square both sides if it is a square root; cube both sides if it is a cube root; etc. It is this step that can introduce extraneous roots if both sides are raised to an even power!!
3. Solve. If the equation still contains radicals, repeat steps 1 and 2. If there are no more radicals, solve the resulting equation.
4. Check for extraneous solutions. Check each solution to confirm the value produces a true statement when substituted back into the original equation.

Example $$\PageIndex{1}$$:

Solve: $$\sqrt { 3 x + 1 } = 4$$.

Solution

\begin{aligned} \sqrt { 3 x + 1 } & = 4 \\ ( \sqrt { 3 x + 1 } ) ^ { 2 } & = ( 4 ) ^ { 2 }\quad\color{Cerulean}{Square \:both\:sides.} \\ 3 x + 1 & = 16 \quad\:\:\:\color{Cerulean}{Solve.}\\ 3 x & = 15 \\ x & = 5 \end{aligned}

Next, we must check.

\begin{aligned} \sqrt { 3 (\color{OliveGreen}{ 5}\color{black}{ )} + 1 } & = 4 \\ \sqrt { 15 + 1 } & = 4 \\ \sqrt { 16 } & = 4 \\ 4 & = 4\:\:\color{Cerulean}{✓} \end{aligned}

The solution set is $$\{5\}$$.

Example $$\PageIndex{2}$$:

Solve $$\sqrt { x - 3 } = x - 5$$.

Solution

\begin{aligned} \sqrt { x - 3 } &= x - 5 \\ ( \sqrt { x - 3 } ) ^ { 2 } &= ( x - 5 ) ^ { 2 }\quad\quad\quad\color{Cerulean}{Square\:both\:sides.} \\ x - 3 &= x ^ { 2 } - 10 x + 25 \end{aligned}

The resulting quadratic equation can be solved by factoring.

\begin{aligned} x - 3 & = x ^ { 2 } - 10 x + 25 \\ 0 & = x ^ { 2 } - 11 x + 28 \\ 0 & = ( x - 4 ) ( x - 7 ) \end{aligned}

$$\begin{array} { r l } { x - 4 = 0 } & { \text { or } \quad x - 7 = 0 } \\ { x = 4 } & \quad\quad\quad\quad\:{ x = 7 } \end{array}$$

Checking the solutions after squaring both sides of an equation is not optional. Use the original equation when performing the check.

 $${\color{Cerulean}{Check}} \text{ } x=4$$ $${\color{Cerulean}{Check}} \text{ } x=7$$ \begin{aligned} \sqrt { x - 3 } & = x - 5 \\ \sqrt { \color{Cerulean}{4}\color{black}{ -} 3 } & = \color{Cerulean}{4}\color{black}{ -} 5 \\ \sqrt { 1 } & = - 1 \\ 1 & = - 1 \quad \color{red}{✗} \end{aligned} \begin{aligned} \sqrt { x - 3 } & = x - 5 \\ \sqrt { \color{Cerulean}{7}\color{black}{ -} 3 } & = \color{Cerulean}{7}\color{black}{ -} 5 \\ \sqrt { 4 } & = 2 \\ 2 & = 2\quad\color{Cerulean}{✓} \end{aligned}

After checking, you can see that $$x = 4$$ is an extraneous solution; it does not solve the original radical equation. Disregard that answer. The solution set consequently is just $$\{ 7 \}$$.

In the previous two examples, notice that the radical is isolated on one side of the equation. Typically, this is not the case. The steps for solving radical equations involving square roots are outlined in the following example.

Example $$\PageIndex{3}$$:

Solve: $$\sqrt { 2 x - 1 } + 2 = x$$.

Solution

Step 1: Isolate the square root.

\begin{aligned} \sqrt { 2 x - 1 } + 2 &= x \\ \sqrt { 2 x - 1 } &= x - 2 \end{aligned}

Step 2: Square both sides.

\begin{aligned} ( \sqrt { 2 x - 1 } ) ^ { 2 } &= ( x - 2 ) ^ { 2 } \\ 2 x - 1 &= x ^ { 2 } - 4 x + 4 \end{aligned}

Step 3: Solve the resulting equation.

\begin{aligned} 2 x - 1 & = x ^ { 2 } - 4 x + 4 \\ 0 & = x ^ { 2 } - 6 x + 5 \\ 0 & = ( x - 1 ) ( x - 5 ) \end{aligned}

$$\begin{array} { r l } { x - 1 = 0 } & { \text { or } \quad x - 5 = 0 } \\ { x = 1 } & \quad\quad\quad\quad{ x = 5 } \end{array}$$

Step 4: Check the solutions in the original equation. Squaring both sides introduces the possibility of extraneous solutions; hence the check is required.

 $${\color{Cerulean}{Check} } \text{ } {x=1}$$ $${\color{Cerulean}{Check} } \text{ } {x=5}$$ \begin{aligned} \sqrt { 2 x - 1 } + 2 & = x \\ \sqrt { 2 ( \color{Cerulean}{1}\color{black}{ )} - 1 } + 2 & = \color{Cerulean}{1} \\ \sqrt { 1 } + 2 & = 1 \\ 1 + 2 & = 1 \\ 3 & = 1 \:\:\color{red}{✗}\end{aligned} \begin{aligned} \sqrt { 2 x - 1 } + 2 & = x \\ \sqrt { 2 ( \color{Cerulean}{5}\color{black}{ )} - 1 } + 2 & = \color{Cerulean}{5} \\ \sqrt { 9 } + 2 & = 5 \\ 3 + 2 & = 5 \\ 5 & = 5\:\:\color{Cerulean}{✓} \end{aligned}

After checking, we can see that $$x = 1$$ is an extraneous solution; it does not solve the original radical equation. This leaves $$\{5 \}$$ as the solution set.

Sometimes there is more than one solution to a radical equation.

Example $$\PageIndex{4}$$:

Solve: $$2 \sqrt { 2 x + 5 } - x = 4$$.

Solution

Begin by isolating the term with the radical.

\begin{aligned} 2 \sqrt { 2 x + 5 } - x &= 4 \quad\quad\color{Cerulean}{Add\:x\:to\:both\:sides.} \\ 2 \sqrt { 2 x + 5 } &= x + 4 \end{aligned}

Despite the fact that the term on the left side has a coefficient, we still consider it to be isolated. Recall that terms are separated by addition or subtraction operators.

\begin{aligned} 2 \sqrt { 2 x + 5 } & = x + 4 \\ ( 2 \sqrt { 2 x + 5 } ) ^ { 2 } & = ( x + 4 ) ^ { 2 } \quad\quad\quad\color{Cerulean}{Square\:both\:sides.}\\ 4 ( 2 x + 5 ) & = x ^ { 2 } + 8 x + 16 \end{aligned}

\begin{aligned} 4 ( 2 x + 5 ) & = x ^ { 2 } + 8 x + 16 \\ 8 x + 20 & = x ^ { 2 } + 8 x + 16 \\ 0 & = x ^ { 2 } - 4 \\ 0 & = ( x + 2 ) ( x - 2 ) \end{aligned}

$$\begin{array} { r l } { x + 2 = 0 } & { \text { or } x - 2 = 0 } \\ { x = - 2 } & \quad\quad\quad\:{ x = 2 } \end{array}$$

Since we squared both sides, we must check our solutions.

 $${\color{Cerulean}{Check} } \text{ } {x=-2}$$ $${\color{Cerulean}{Check} } \text{ }{x=2}$$ $$\begin{array} { r } { 2 \sqrt { 2 x + 5 } - x = 4 } \\ { 2 \sqrt { 2 ( \color{Cerulean}{- 2}\color{black}{ )} + 5 } - ( \color{Cerulean}{- 2}\color{black}{ )} = 4 } \\ { 2 \sqrt { - 4 + 5 } + 2 = 4 } \\ { 2 \sqrt { 1 } + 2 = 4 } \\ { 2 + 2 = 4 } \\ { 4 = 4 }\:\:\color{Cerulean}{✓} \end{array}$$ \begin{aligned} 2 \sqrt { 2 x + 5 } - x & = 4 \\ 2 \sqrt { 2 (\color{Cerulean}{ 2}\color{black}{ )} + 5 } - ( \color{Cerulean}{2}\color{black}{ )} & = 4 \\ 2 \sqrt { 4 + 5 } - 2 & = 4 \\ 2 \sqrt { 9 } - 2 & = 4 \\ 6 - 2 & = 4 \\ 4 & = 4 \:\:\color{Cerulean}{✓}\end{aligned}

After checking, we can see that both are solutions to the original equation. The solution set is $$\{ \pm 2 \}$$.

Sometimes both of the possible solutions are extraneous.

Example $$\PageIndex{5}$$:

Solve: $$\sqrt { 4 - 11 x } - x + 2 = 0$$.

Solution

\begin{aligned} \sqrt { 4 - 11 x } - x + 2 & = 0\quad\quad\quad\quad\quad\color{Cerulean}{Isolate\:the\:radical.} \\ \sqrt { 4 - 11 x } & = x - 2 \\ ( \sqrt { 4 - 11 x } ) ^ { 2 } & = ( x - 2 ) ^ { 2 }\quad\quad\color{Cerulean}{Square\:both\:sides.} \\ 4 - 11 x & = x ^ { 2 } - 4 x + 4\:\:\color{Cerulean}{Solve.} \\ 0 & = x ^ { 2 } + 7 x \\ 0 & = x ( x + 7 ) \end{aligned}

\begin{aligned} x = 0 \text { or } x + 7 & = 0 \\ x & = - 7 \end{aligned}

Since we squared both sides, we must check our solutions.

 $${\color{Cerulean}{Check } } \text{ } {x=0}$$ $${\color{Cerulean}{Check } } \text{ } {x=-7}$$ \begin{aligned} \sqrt { 4 - 11 x } - x + 2 & = 0 \\ \sqrt { 4 - 11 ( \color{Cerulean}{0}\color{black}{ )} } -\color{Cerulean}{ 0}\color{black}{ +} 2 & = 0 \\ \sqrt { 4 } + 2 & = 0 \\ 2 + 2 & = 0 \\ 4 & = 0 \:\:\color{red}{✗} \end{aligned} \begin{aligned} \sqrt { 4 - 11 x } - x + 2 &=0 \\ \sqrt { 4 - 11 ( \color{Cerulean}{- 7}\color{black}{ )} } - ( \color{Cerulean}{- 7}\color{black}{ )} + 2 &=0 \\ \sqrt { 4 + 77 } + 7 + 2 &=0 \\ \sqrt { 81 } + 9 &=0 \\ 9 + 9 &=0 \\ 18 &=0 \:\:\color{red}{✗} \end{aligned}

Since both possible solutions are extraneous, the equation has no solution and the solution set is $$\{ \:\: \}$$.

The squaring property of equality extends to any positive integer power $$n$$. Given real numbers $$a$$ and $$b$$, the power property of equality states: $$\text{If}\:\:\:a = b , \text { then } a ^ { n } = b ^ { n }$$. This, and the fact that $$( \sqrt [ n ] { a } ) ^ { n } = \sqrt [ n ] { a ^ { n } } = a$$, when $$a$$ is nonnegative, is used to solve radical equations with indices greater than $$2$$.

Example $$\PageIndex{6}$$:

Solve $$\sqrt [ 3 ] { 4 x ^ { 2 } + 7 } - 2 = 0$$.

Solution

\begin{aligned} \sqrt [ 3 ] { 4 x ^ { 2 } + 7 } - 2 & = 0\quad\quad\color{Cerulean}{Isolate\:the\:radical.} \\ \sqrt [ 3 ] { 4 x ^ { 2 } + 7 } & = 2 \\ \left( \sqrt [ 3 ] { 4 x ^ { 2 } + 7 } \right) ^ { 3 } & = ( 2 ) ^ { 3 }\quad\color{Cerulean}{Cube\:both\:sides.} \\ 4 x ^ { 2 } + 7 & = 8 \quad\quad\color{Cerulean}{Solve.}\\ 4 x ^ { 2 } - 1 & = 0 \\ ( 2 x + 1 ) ( 2 x - 1 ) & = 0 \end{aligned}

$$\begin{array} { r l } { 2 x + 1 = 0 } & { \text { or } \quad 2 x - 1 = 0 } \\ { 2 x = - 1 } &\quad\quad\quad\quad\: { 2 x = 1 } \\ { x = - \frac { 1 } { 2 } } &\quad\quad\quad\quad\:\:\; { x = \frac { 1 } { 2 } } \end{array}$$

 $${\color{Cerulean}{Check} } \text{ } {x=-\frac{1}{2}}$$ $${\color{Cerulean}{Check} } \text{ } {x=\frac{1}{2}}$$ \begin{aligned} \sqrt [ 3 ] { 4 x ^ { 2 } + 7 } - 2 & = 0 \\ \sqrt [ 3 ] { 4 \left( \color{Cerulean}{- \frac { 1 } { 2} } \right) ^ { 2 } + 7 } - 2 & = 0 \\ \sqrt [ 3 ] { 4 \cdot \frac { 1 } { 4 } + 7 } - 2 & = 0 \\ \sqrt [ 3 ] { 8 } - 2 & = 0 \\ 2- 2 & = 0 \\ 0 & = 0\:\:\color{Cerulean}{✓} \end{aligned} \begin{aligned} \sqrt [ 3 ] { 4 x ^ { 2 } + 7 } - 2 & = 0 \\ \sqrt [ 3 ] { 4 \left( \color{Cerulean}{\frac { 1 } { 2} } \right) ^ { 2 } + 7 } - 2 & = 0 \\ \sqrt [ 3 ] { 4 \cdot \frac { 1 } { 4 } + 7 } - 2 & = 0 \\ \sqrt[3]{1+7}-2 &=0 \\ \sqrt [ 3 ] { 8 } - 2 & = 0 \\ 2 - 2 & = 0 \\ 0 & = 0\:\:\color{Cerulean}{✓} \end{aligned}

The solution set is  $$\Large\{ \pm \frac { 1 } { 2 } \Large\}$$.

Try It $$\PageIndex{6}$$

Solve: $$x - 3 \sqrt { 3 x + 1 } = 3$$

The solution is $$33$$. (The other proposed solution, $$x=0$$ was rejected. )

It may be the case that the equation has more than one term that consists of radical expressions.

Example $$\PageIndex{7}$$:

Solve: $$\sqrt { 5 x - 3 } = \sqrt { 4 x - 1 }$$.

Solution

Both radicals are considered isolated on separate sides of the equation.

\begin{aligned} \sqrt { 5 x - 3 } & = \sqrt { 4 x - 1 } \\ ( \sqrt { 5 x - 3 } ) ^ { 2 } & = ( \sqrt { 4 x - 1 } ) ^ { 2 } \quad\color{Cerulean}{Square\:both\:sides.}\\ 5 x - 3 & = 4 x - 1 \quad\quad\quad\color{Cerulean}{Solve.}\\ x & = 2 \end{aligned}

Check $$x=2$$.

\begin{aligned} \sqrt { 5 x - 3 } & = \sqrt { 4 x - 1 } \\ \sqrt { 5 ( \color{OliveGreen}{2}\color{black}{ )} - 3 } & = \sqrt { 4 ( \color{OliveGreen}{2}\color{black}{ )} - 1 } \\ \sqrt { 10 - 3 } & = \sqrt { 8 - 1 } \\ \sqrt { 7 } & = \sqrt { 7 }\:\:\color{Cerulean}{✓} \end{aligned}

The solution set is $$\{ 2 \}$$.

Example $$\PageIndex{8}$$:

Solve: $$\sqrt [ 3 ] { x ^ { 2 } + x - 14 } = \sqrt [ 3 ] { x + 50 }$$.

Solution

Eliminate the radicals by cubing both sides.

\begin{aligned} \sqrt [ 3 ] { x ^ { 2 } + x - 14 } & = \sqrt [ 3 ] { x + 50 } \\ \left( \sqrt [ 3 ] { x ^ { 2 } + x - 14 } \right) ^ { 3 } & = ( \sqrt [ 3 ] { x + 50 } ) ^ { 3 }\quad\color{Cerulean}{Cube\:both\:sides.} \\ x ^ { 2 } + x - 14 & = x + 50 \quad\quad\quad\color{Cerulean}{Solve.}\\ x ^ { 2 } - 64 & = 0 \\ ( x + 8 ) ( x - 8 ) & = 0 \end{aligned}

$$\begin{array} { r l } { x + 8 = 0 } & { \text { or } \quad x - 8 = 0 } \\ { x = - 8 } & \quad\quad\quad\quad{ x = 8 } \end{array}$$

 $${\color{Cerulean}{Check} } \text{ } {x=-8}$$ $${\color{Cerulean}{Check}} \text{ } {x=8}$$ \begin{aligned} \sqrt [ 3 ] { x ^ { 2 } + x - 14 } & = \sqrt [ 3 ] { x + 50 } \\ \sqrt [ 3 ] { ( \color{Cerulean}{- 8}\color{black}{ )} ^ { 2 } + ( \color{Cerulean}{- 8}\color{black}{ )} - 14 } & = \sqrt [ 3 ] { ( \color{Cerulean}{- 8}\color{black}{ )} + 50 } \\ \sqrt [ 3 ] { 64 - 8 - 14 } & = \sqrt [ 3 ] { 42 } \\ \sqrt [ 3 ] { 42 } & = \sqrt [ 3 ] { 42 }\:\:\color{Cerulean}{✓} \end{aligned} \begin{aligned} \sqrt [ 3 ] { x ^ { 2 } + x - 14 } & = \sqrt [ 3 ] { x + 50 } \\ \sqrt [ 3 ] { ( \color{Cerulean}{ 8}\color{black}{ )} ^ { 2 } + ( \color{Cerulean}{ 8}\color{black}{ )} - 14 } & = \sqrt [ 3 ] { ( \color{Cerulean}{8}\color{black}{ )} + 50 } \\ \sqrt [ 3 ] { 64 + 8 - 14 } & = \sqrt [ 3 ] { 58 } \\ \sqrt [ 3 ] { 58 } & = \sqrt [ 3 ] { 58 }\:\:\color{Cerulean}{✓} \end{aligned}

The solution set is $$\{ \pm 8 \}$$.

It may not be possible to isolate a radical on both sides of the equation. When this is the case, isolate the radicals, one at a time, and apply the squaring property of equality multiple times until only a polynomial remains.

Example $$\PageIndex{9}$$:

Solve: $$\sqrt { x + 2 } - \sqrt { x } = 1$$

Solution

\begin{aligned} \sqrt { x + 2 } - \sqrt { x } & = 1 \\ \sqrt { x + 2 } & = \sqrt { x } + 1 \end{aligned}

Square both sides. Be careful to apply the distributive property correctly to the right side.

\begin{aligned} ( \sqrt { x + 2 } ) ^ { 2 } & = ( \sqrt { x } + 1 ) ^ { 2 } \\ x + 2 & = ( \sqrt { x } + 1 ) ( \sqrt { x } + 1 ) \\ x + 2 & = \sqrt { x ^ { 2 } } + \sqrt { x } + \sqrt { x } + 1 \\ x + 2 & = x + 2 \sqrt { x } + 1 \end{aligned}

Now the equation contains only one radical. Isolate it and square both sides again.

\begin{aligned} x + 2 & = x + 2 \sqrt { x } + 1 \\ 1 & = 2 \sqrt { x } \\ ( 1 ) ^ { 2 } & = ( 2 \sqrt { x } ) ^ { 2 } \\ 1 & = 4 x \\ \frac { 1 } { 4 } & = x \end{aligned}

Check to see if $$x = \frac { 1 } { 4 }$$ satisfies the original equation $$\sqrt { x + 2 } - \sqrt { x } = 1$$

$$\begin{array} { r } { \sqrt { \color{OliveGreen}{\frac { 1 } { 4 }}\color{black}{ +} 2 } - \sqrt { \color{OliveGreen}{\frac { 1 } { 4 }} } \color{black}{=} 1 } \\ { \sqrt { \frac { 9 } { 4 } } - \frac { 1 } { 2 } = 1 } \\ { \frac { 3 } { 2 } - \frac { 1 } { 2 } = 1 } \\ { \frac { 2 } { 2 } = 1 } \\ { 1 = 1 }\color{Cerulean}{✓} \end{array}$$

The solution set is  $$\large\{ \frac { 1 } { 4 } \large\}$$.

Note

Observe that $$( A + B ) ^ { 2 } \neq A ^ { 2 } + B ^ { 2 }$$, even though $$( A \cdot B ) ^ { 2 } = A ^ { 2 } \cdot B ^ { 2 }$$!!

For example,

$$\color{Cerulean}{(}\color{black}{ \sqrt { x + 2 }}\color{Cerulean}{ ) ^ { 2 } }\color{black}{-}\color{Cerulean}{ (}\color{black}{ \sqrt { x }}\color{Cerulean}{ ) ^ { 2 }}\color{black}{ =}\color{Cerulean}{ (}\color{black}{ 1}\color{Cerulean}{ ) ^ { 2 }}\\\color{red}{Incorrect!}$$

This is a common mistake and leads to an incorrect result. When squaring both sides of an equation with multiple terms, we must take care to apply the distributive property correctly.

Example $$\PageIndex{10}$$:

Solve: $$\sqrt { 2 x + 10 } - \sqrt { x + 6 } = 1$$

Solution

\begin{aligned} \sqrt { 2 x + 10 } - \sqrt { x + 6 } &= 1 \\ \sqrt { 2 x + 10 } & = \sqrt { x + 6 } + 1 \end{aligned}

Square both sides. Take care to apply the distributive property CORRECTLY to the right side.

\begin{aligned} ( \sqrt { 2 x + 10 } ) ^ { 2 } & = ( \sqrt { x + 6 } + 1 ) ^ { 2 } \\ 2 x + 10 & = x + 6 + 2 \sqrt { x + 6 } + 1 \\ 2 x + 10 & = x + 7 + 2 \sqrt { x + 6 } \end{aligned}

At this point we have one term that contains a radical. Isolate it and square both sides again.

\begin{aligned} 2 x + 10 & = x + 7 + 2 \sqrt { x + 6 } \\ x + 3 & = 2 \sqrt { x + 6 } \\ ( x + 3 ) ^ { 2 } & = ( 2 \sqrt { x + 6 } ) ^ { 2 } \\ x ^ { 2 } + 6 x + 9 & = 4 ( x + 6 ) \\ x ^ { 2 } + 6 x + 9 & = 4 x + 24 \\ x ^ { 2 } + 2 x - 15 & = 0 \\ (x - 3 ) ( x + 5 ) & = 0 \end{aligned}

$$\begin{array} { r l } { x - 3 = 0 } & { \text { or } \quad x + 5 = 0 } \\ { x = 3 } & \quad\quad\quad\quad\:{ x = - 5 } \end{array}$$

 $${\color{Cerulean}{Check } } \text{ } {x=3}$$ $${\color{Cerulean}{Check } } \text{ } {x=-5}$$ \begin{aligned} \sqrt { 2 x + 10 } - \sqrt { x + 6 } & = 1 \\ \sqrt { 2 ( \color{Cerulean}{3}\color{black}{ )} + 10 } - \sqrt { \color{Cerulean}{3}\color{black}{ +} 6 } & = 1 \\ \sqrt { 16 } - \sqrt { 9 } & = 1 \\ 4 - 3 & = 1 \\ 1 & = 1\:\:\color{Cerulean}{✓} \end{aligned} \begin{aligned} \sqrt { 2 x + 10 } - \sqrt { x + 6 } &= 1 \\ \sqrt { 2 (\color{Cerulean}{ - 5}\color{black}{ )} + 10 } - \sqrt { \color{Cerulean}{- 5}\color{black}{ +} 6 } &=1 \\ \sqrt { 0 } - \sqrt { 1 }& =1 \\ 0 - 1 &=1 \\ - 1 &=1\:\:\color{red}{✗} \end{aligned}

The solution set is $$\{3\}$$.

Try It $$\PageIndex{10}$$

Solve: $$\sqrt { 4 x + 21 } - \sqrt { 2 x + 22 } = 1$$

The solution is $$7$$  (The other proposed solution, $$x=-3$$ was rejected. )