
To solve an equation of form $a\sqrt[n]{u}=b\text{,}$ we

1. divide both sides of the equation by $a$ if that doesn't introduce an unnecessary fraction
2. raise both sides of the equation to the $n^{th}$ power; make sure that you remember to apply the exponent to the factor of $a$
3. solve the resultant equation
4. check each potential solution in the original equation and state your conclusion.

Unlike solving other types of equations, the last step (checking) is not "optional." Raising both sides of an equation to the $n^{th}$ power can result in what we call "extraneous solutions." Extraneous solutions are solutions to the new equation that are not also solutions to the original equation. This can happen because, for example, squaring both sides of the false equation $-2=2$ results in the true equation $4=4\text{.}$

###### Example11.9.1

Determine the solution set to the equation $\sqrt[4]{9-2x}=3\text{.}$

Solution

We begin by raising both sides of the equation to the fourth power and then we solve the resultant equation.

\begin{align*} \sqrt[4]{9-2x}\amp=3\\ \highlight{(}\sqrt[4]{9-2x}\highlight{)^4}\amp=\highlight{(}3\highlight{)^4}\\ 9-2x\amp=81\\ 9-2x\subtractright{9}\amp=81\subtractright{9}\\ -2x\amp=72\\ \divideunder{-2x}{-2}\amp=\divideunder{72}{-2}\\ x\amp=-36 \end{align*}

We now check the solution of $-36$ in the original equation.

\begin{align*} \sqrt[4]{9-2 \cdot \highlight{-36}}\amp=3\text{ ?}\\ \sqrt[4]{9+72}\amp=81\text{ ?}\\ \sqrt[4]{81}\amp=3\,\checkmark \end{align*}

The solution set to the equation $\sqrt[4]{9-2x}=3$ is $\{-36\}\text{.}$

###### Example11.9.2

Determine the solution set to the equation $\sqrt{x+5}=-7\text{.}$

Solution

We begin by squaring both sides of the equation and then we solve the resultant equation.

\begin{align*} \sqrt{x+5}\amp=-7\\ \highlight{(}\sqrt{x+5}\highlight{)^4}\amp=\highlight{(}-7\highlight{)^4}\\ x+5\amp=49\\ x+5\subtractright{5}\amp=49\subtractright{5}\\ x\amp=42 \end{align*}

When we check the solution we arrive at the equation

\begin{equation*} \sqrt{49}=-7 \end{equation*}

which is false, by definition $\sqrt{x}$ always represents a non-negative value. If we want to reference the negative square-root of $49\text{,}$ we need to write $-\sqrt{49}\text{.}$ So the solution set to the given equation is $\emptyset\text{.}$

You may think that the last example was contrived, and you would be correct in that thought. However, in the following sections the equations won't be so blatantly contrived, and we need to get it established straight away that the check for extraneous solutions must be done each and every time.

###### Example11.9.3

Determine the solution set to the equation $3\sqrt{\frac{x+7}{2}}=5\text{.}$

Solution

We would begin by dividing both sides of the equation by $3\text{,}$ but doing so would introduce an unnecessary fraction to the right-side of the equation. I think we'd be better off just squaring both sides of the equation as is. We need to be mindful, however, that both factors on the left-side of the equation need to be squared.

\begin{align*} 3\sqrt{\frac{x+7}{2}}\amp=5\\ \left(3\sqrt{\frac{x+7}{2}}\right)^2\amp=5^2\\ 9 \cdot \frac{x+7}{2}\amp=25\\ \frac{9x+63}{2}\amp=25\\ \multiplyleft{2}\frac{9x+63}{2}\amp=\multiplyleft{2}25\text{ (clearing the fraction)}\\ 9x+63\amp=50\\ 9x+63\subtractright{63}\amp=50\subtractright{63}\\ 9x\amp=-13\\ \divideunder{9x}{9}\amp=\divideunder{-13}{9}\\ x\amp=-\frac{13}{9} \end{align*}

Let's check our solution.

\begin{align*} 3\sqrt{\frac{\highlight{-\frac{13}{9}}+7}{2}}\amp=5\text{ ?}\\ 3\sqrt{\frac{-\frac{13}{9}+\frac{63}{9}}{2}}\amp=5\text{ ?}\\ 3\sqrt{\frac{\frac{50}{9}}{\frac{2}{1}}}\amp=5\text{ ?}\\ 3\sqrt{\frac{50}{9} \cdot \frac{1}{2}}\amp=5\text{ ?}\\ 3\sqrt{\frac{25}{9}}\amp=5\text{ ?}\\ 3 \cdot \frac{5}{3}\amp=5\,\checkmark \end{align*}

The solution set is $\{-\frac{13}{9}\}\text{.}$

###### Example11.9.4

Determine the solution set to the equation $4\sqrt[5]{\frac{2x}{3}-6}=-8$

Solution

This time we will begin by dividing both sides of the equation by $4\text{.}$ This won't introduce any unnecessary fractions and it will set us up for much smaller values when we raise both sides of the equation to the fifth power.

Let's check the value of $-39$ in the original equation.

\begin{align*} 4\sqrt[5]{\frac{2 \cdot \highlight{-39}}{3}-6}\amp=-8\text{ ?}\\ 4\sqrt[5]{-32}\amp=-8\text{ ?}\\ 4 \cdot -2\amp=-8\,\checkmark \end{align*}

The solution set to the stated equation is $\{-39\}\text{.}$

When presented with an equation that contains a single radical expression (or repeated occurrences of like radical expressions), our first objective is to isolate the radical expression, $a\sqrt[n]{u}\text{,}$ and then raise both sides of the equation to $n^{th}$ power. We solve the resultant equation and make sure that we check all potential solutions in the original equation. Of course, we then satte our conclusion. Let's see a few examples.

###### Example11.9.5

Determine the solution set to the equation $\frac{\sqrt{18+2x}}{3}-7=-4\text{.}$

Solution

We begin by adding $7$ to both sides of the equation and then clearing the fraction. We then square both sides of the equation and solve the resultant equation.

\begin{align*} \frac{\sqrt{18+2x}}{3}-7\amp=-4\\ \frac{\sqrt{18+2x}}{3}-7\addright{7}\amp=-4\addright{7}\\ \frac{\sqrt{18+2x}}{3}\amp=3\\ \multiplyleft{3}\frac{\sqrt{18+2x}}{3}\amp=\multiplyleft{3}3\\ \sqrt{18+2x}\amp=9\\ \highlight{(}\sqrt{18+2x}\highlight{)^2}\amp=\highlight{(}9\highlight{)^2}\\ 18+2x\amp=81\\ 18+2x\subtractright{18}\amp=81\subtractright{18}\\ 2x\amp=63\\ \divideunder{2x}{2}\amp=\divideunder{63}{2}\\ x\amp=\frac{63}{2} \end{align*}

We now check the potential solution in the original equation.

\begin{align*} \frac{\sqrt{18+2 \cdot \highlight{\frac{63}{2}}}}{3}-7\amp=-4\text{ ?}\\ \frac{\sqrt{18+63}}{3}-7\amp=-4\text{ ?}\\ \frac{\sqrt{81}}{3}-7\amp=4\text{ ?}\\ \frac{9}{3}-7\amp=-4\text{ ?}\\ 3-7\amp=-4\,\checkmark \end{align*}

The solution set to the given equation is $\frac{63}{2}\text{.}$

###### Example11.9.6

Determine the solution set to the equation $2-3\sqrt[3]{4-5x}=7+2\sqrt[3]{4-5x}\text{.}$

Solution

Because the radical expressions are like terms, we want to combine them with the constant term on the other side of the equal sign. We then cube both sides of the equation and proceed to solve the resultant equation.

\begin{align*} 2-3\sqrt[3]{4-5x}\amp=7+2\sqrt[3]{4-5x}\\ 2-3\sqrt[3]{4-5x}\addright{\sqrt[3]{4-5x}}\subtractright{7}\amp=7+2\sqrt[3]{4-5x}\addright{\sqrt[3]{4-5x}}\subtractright{7}\\ -5\amp=5\sqrt[3]{4-5x}\\ \divideunder{-5}{5}\amp=\divideunder{5\sqrt[3]{4-5x}}{5}\\ -1\amp=\sqrt[3]{4-5x}\\ \highlight{(}-1\highlight{)^3}\amp=\highlight{(}\sqrt[3]{4-5x}\highlight{)^3}\\ -1\amp=4-5x\\ -1\subtractright{4}\amp=4-5x\subtractright{4}\\ -5\amp=-5x\\ \divideunder{-5}{-5}\amp=\divideunder{-5x}{-5}\\ 1\amp=x \end{align*}

We now check our potential solution in the original equation.

\begin{align*} 2-3\sqrt[3]{4-5 \cdot \highlight{1}}\amp=7+2\sqrt[3]{4-5 \cdot \highlight{1}}\text{ ?}\\ 2-3\sqrt[3]{-1}\amp=7+2\sqrt[3]{-1}\text{ ?}\\ 2-3 \cdot -1\amp=7+2 \cdot -1\text{ ?}\\ 2+3\amp=7-2\text{ ?}\\ 5\amp=5\,\checkmark \end{align*}

The solution set is $\{1\}\text{.}$

###### Example11.9.7

Determine the solution set to the equation $3y-\sqrt{y}-2=0\text{.}$

Solution

We begin by isolating the square root expression and squaring both sides of the equation. We then solve the resultant equation.

\begin{align*} 3y+\sqrt{y}-2\amp=0\\ 3y+\sqrt{y}-2\subtractright{3y}\addright{2}\amp=0\subtractright{3y}\addright{2}\\ \sqrt{y}\amp=-3y+2\\ \highlight{(}\sqrt{y}\highlight{)^2}\amp=\highlight{(}-3y+2\highlight{)^2}\\ y\amp=9y^2-12y+4\\ y\subtractright{y}\amp=9y^2-12y+4\subtractright{y}\\ 0\amp=9y^2-13y+4\\ 0\amp=9y^2-9y-4y+4\\ 0\amp=9y\highlightr{(y-1)}-4\highlightr{(y-1)}\\ 0\amp=(9y-4)\highlightr{(y-1)} \end{align*}
\begin{align*} 9x-4\amp=0 \amp\amp\text{ or }\amp y-1\amp=0\\ 9x-4\addright{4}\amp=0\addright{4} \amp\amp\text{ or }\amp y-1\addright{1}\amp=0\addright{1}\\ 9x\amp=4 \amp\amp\text{ or }\amp y\amp=1\\ \divideunder{9x}{9}\amp=\divideunder{4}{9} \amp\amp\text{ or }\amp y\amp=1\\ y\amp=\frac{4}{9} \amp\amp\text{ or }\amp y\amp=1 \end{align*}

We now check each potential solution in the original equation.

\begin{align*} 3 \cdot \highlight{\frac{4}{9}}+\sqrt{\highlight{\frac{4}{9}}}-2\amp=0\text{ ?}\\ \frac{4}{3}+\frac{2}{3}-2\amp=0\text{ ?}\\ 0\amp=0\,\checkmark \end{align*}
\begin{align*} 3 \cdot \highlight{1}+\sqrt{\highlight{1}}-2\amp=0\text{ ?}\\ 3 \cdot 1 +1-2\amp=0\text{ ?}\\ 2\amp=0 \text{ false} \end{align*}

The solution set is $\{\frac{4}{9}\}\text{.}$

In the previous examples, whenever two occurrences of radical terms occurred they were always like radicals. In those examples we were able to combine the radicals. When the radical expressions are not like terms, that strategy is not viable. In this case we restrict ourselves to equations with two unlike square root expressions. The problem solving strategy follows.

1. Isolate one of the square root expressions and square both sides of the equation.
2. If there is still a square root expression in the equation, isolate it and square both sides of the equation once again.
3. Regardless of whether or not you had to execute step $2\text{,}$ solve the resultant equation.
4. Check each potential solution in the original equation and state your conclusion.

Let's see a few examples.

###### Example11.9.8

Determine the solution set to the equation $\sqrt{1-7x}-2\sqrt{-6-3x}=0$

Solution

Because there are no terms that occur outside of the two square root terms, we will only have to square both sides of the equation once, after first isolating the square root expressions on opposite sides of the equal sign.

Let's check our potential solution in our original equation.

\begin{align*} \sqrt{1-7 \cdot \highlight{-5}}-2\sqrt{-6-3 \cdot \highlight{-5}}\amp=0\text{ ?}\\ \sqrt{36}-2\sqrt{9}\amp=0\text{ ?}\\ 6-2 \cdot 3\amp=0\text{ ?}\\ 6-6\amp=0\,\checkmark \end{align*}

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

###### Example11.9.9

Determine the solution set to the equation $\sqrt{x}-\sqrt{x-7}=1\text{.}$

Solution

We begin by isolating $\sqrt{x}$ and squaring both sides of the equation. Because of the term of $1$ that occurs outside of a square root, we need to FOIL that side of the equation, and the result will contain yet another square root term. We will then isolate that square root expression, square both sides again, and proceed as normal.

Let's check our potential solution in the original equation.

\begin{align*} \sqrt{\highlight{16}}-\sqrt{\highlight{16}-7}\amp=1\text{ ?}\\ \sqrt{16}-\sqrt{9}\amp=1\text{ ?}\\ 4-3\amp=a\,\checkmark \end{align*}

The solution set is $\{16\}\text{.}$

###### Example11.9.10

Determine the solution set to the equation $\sqrt{2x-3}-\sqrt{4x+2}=-2\text{.}$

Solution

We begin by isolating $\sqrt{2x-3}$ and squaring both sides of the equation. Because of the term of $6$ that occurs outside of a square root, we need to FOIL that side of the equation, and the result will contain yet another square root term. We will then isolate that square root expression, square both sides again, and proceed as normal.

\begin{align*} \sqrt{2x-3}-\sqrt{4x+2}\amp=-2\\ \sqrt{2x-3}-\sqrt{4x+2}\addright{\sqrt{4x+2}}\amp=-2\addright{\sqrt{4x+2}}\\ \sqrt{2x-3}\amp=\sqrt{4x+2}-2\\ \highlight{(}\sqrt{2x-3}\highlight{)^2}\amp=\highlight{(}\sqrt{4x+2}-2\highlight{)^2}\\ 2x-3\amp=4x+2-4\sqrt{4x+2}+4\\ 2x-3\amp=4x+6-4\sqrt{4x+2}\\ 2x-3\subtractright{4x}\subtractright{6}\amp=4x+6-4\sqrt{4x+2}\subtractright{4x}\subtractright{6}\\ -2x-9\amp=-4\sqrt{4x+2}\\ \highlight{(}-2x-9\highlight{)^2}\amp=\highlight{(}-4\sqrt{4x+2}\highlight{)^2}\\ 4x^2+36x+81\amp=16(4x+2)\\ 4x^2+36x+81\amp=64x+32\\ 4x^2+36x+81\subtractright{64x}\subtractright{32}\amp=64x+32\subtractright{64x}\subtractright{32}\\ 4x^2-28x+49\amp=0\\ 4x^2-14x-14x+49\amp=0\\ 2x\highlightr{(2x-7)}-7\highlightr{(2x-7)}\amp=0\\ (2x-7)\highlightr{(2x-7)}\amp=0\\ 2x-7\amp=0\\ 2x-7\addright{7}\amp=0\addright{7}\\ 2x\amp=7\\ \divideunder{2x}{2}\amp=\divideunder{7}{2}\\ x\amp=\frac{7}{2} \end{align*}

Let's check our potential solution in the original equation.

\begin{align*} \sqrt{2 \cdot \highlight{\frac{7}{2}}-3}-\sqrt{4 \cdot \highlight{\frac{7}{2}}+2}\amp=-2\text{ ?}\\ \sqrt{4}-\sqrt{16}\amp-2\text{ ?}\\ 2-4\amp=-2\,\checkmark \end{align*}

The solution set is $\{\frac{7}{2}\}\text{.}$

### Subsection11.9.1Exercises

Determine the solution set to each equation.

###### 1

$\sqrt[5]{2x-1}=3$

Solution

We begin by raising both sides of the equation to the fifth power. We then solve the resultant equation.

We now check the potential solution in the original equation.

\begin{align*} \sqrt[5]{2 \cdot \highlight{122}-1}\amp=3\text{ ?}\\ \sqrt[5]{243}\amp=3\,\checkmark \end{align*}

The solution set is $\{122\}\text{.}$

###### 2

$4\sqrt{2-x}=5$

Solution

We begin by squaring both sides of the equation. We then solve the resultant equation.

\begin{align*} 4\sqrt{2-x}\amp=5\\ \highlight{(}4\sqrt{2-x}\highlight{)^2}\amp=\highlight{(}5\highlight{)^2}\\ 16(2-x)\amp=25\\ 32-16x\amp=25\\ 32-16x\subtractright{32}\amp=25\subtractright{32}\\ -16x\amp=-7\\ \divideunder{-16x}{-16}\amp=\divideunder{-7}{-16}\\ x\amp=\frac{7}{16} \end{align*}

We now check the potential solution in the original equation.

\begin{align*} 4\sqrt{2-\highlight{\frac{7}{16}}}\amp=5\text{ ?}\\ 4\sqrt{\frac{25}{16}}\amp=5\text{ ?}\\ 4 \cdot \frac{5}{4}\amp=5\,\checkmark \end{align*}

The solution set is $\{\frac{7}{16}\}$

###### 3

$\sqrt[3]{\frac{x}{4}+2}=-4$

Solution

We begin by raising both sides of the equation to the third power. We then solve the resultant equation.

\begin{align*} \sqrt[3]{\frac{x}{4}+2}\amp=-4\\ \left(\sqrt[3]{\frac{x}{4}+2}\right)^3\amp=(-4)^3\\ \frac{x}{4}+2\amp=-64\\ \frac{x}{4}+2\subtractright{2}\amp=-64\subtractright{2}\\ \frac{x}{4}\amp=-66\\ \multiplyleft{4}\frac{x}{4}\amp=\multiplyleft{4}-66\\ x\amp=-264 \end{align*}

We now check the potential solution in the original equation.

\begin{align*} \sqrt[3]{\frac{\highlight{-264}}{4}+2}\amp=-4\text{ ?}\\ \sqrt[3]{-66+2}\amp=-4\text{ ?}\\ \sqrt[3]{-64}\amp=-4\,\checkmark \end{align*}

The solution set is $\{-264\}\text{.}$

###### 4

$3\sqrt[4]{x+9}=3$

Solution

We begin by dividing three from both sides of the equation and then raising both sides of the equation to the fourth power. We then solve the resultant equation.

\begin{align*} 3\sqrt[4]{x+9}\amp=3\\ \divideunder{3\sqrt[4]{x+9}}{3}\amp=\divideunder{3}{3}\\ \sqrt[4]{x+9}\amp=1\\ \highlight{(}\sqrt[4]{x+9}\highlight{)^4}\amp=\highlight{(}1\highlight{)^4}\\ x+9\amp=1\\ x+9\subtractright{9}\amp=1\subtractright{9}\\ x\amp=-8 \end{align*}

We now check the potential solution in the original equation.

\begin{align*} 3\sqrt[4]{\highlight{-8}+9}\amp=3\text{ ?}\\ 3\sqrt{1}\amp=3\text{ ?}\\ 3 \cdot 1\amp=3\,\checkmark \end{align*}

The solution set is $\{-8\}\text{.}$

###### 5

$3\sqrt{2y+7}-2=7$

Solution

We begin by isolating the square root expression. We then square both sides of the equation and solve the resultant equation.

\begin{align*} 3\sqrt{2y+7}-2\amp=7\\ 3\sqrt{2y+7}-2\addright{2}\amp=7\addright{2}\\ 3\sqrt{2y+7}\amp=9\\ \divideunder{3\sqrt{2y+7}}{3}\amp=\divideunder{9}{3}\\ \sqrt{2y+7}\amp=3\\ \highlight{(}\sqrt{2y+7}\highlight{)^2}\amp=\highlight{(}3\highlight{)^2}\\ 2y+7\amp=9\\ 2y+7\subtractright{7}\amp=9\subtractright{7}\\ 2y\amp=2\\ \divideunder{2y}{2}\amp=\divideunder{2}{2}\\ y\amp=1 \end{align*}

We now check our potential solution.

\begin{align*} 3\sqrt{2 \cdot \highlight{1}+7}-2\amp=7\text{ ?}\\ 3\sqrt{9}-2\amp=7\text{ ?}\\ 3 \cdot 3-2\amp=7\text{ ?}\\ 9-2\amp=7\,\checkmark \end{align*}

The solution set is $\{1\}$

###### 6

$5-\sqrt[4]{11-x}=8$

Solution

We begin by isolating the fourth root expression. We then raise both sides of the equation to the fourth power and solve the resultant equation.

\begin{align*} 5-\sqrt[4]{11-x}\amp=8\\ 5-\sqrt[4]{11-x}\subtractright{5}\amp=8\subtractright{5}\\ -\sqrt[4]{11-x}\amp=3\\ \highlight{(}-\sqrt[4]{11-x}\highlight{)^4}\amp=\highlight{(}3\highlight{)^4}\\ 11-x\amp=81\\ 11-x\subtractright{11}\amp=81\subtractright{11}\\ -x\amp=70\\ \multiplyleft{-1}-x\amp=\multiplyleft{-1}70\\ x\amp=-70 \end{align*}

We now check our potential solution.

\begin{align*} 5-\sqrt[4]{11-(\highlight{-70})}\amp=8\text{ ?}\\ 5-\sqrt[4]{81}\amp=8\text{ ?}\\ 5-3\amp=8\text{ ?}\\ 2\amp=8\text{ no siree!} \end{align*}

Since our only potential solution turned out to be extraneous, the solution set is $\emptyset\text{.}$

###### 7

$\sqrt[3]{x+17}-\sqrt[3]{4x-13}=0$

Solution

We begin by isolating the cube root expressions on opposite sides of the equal sign. We then cube both sides of the equation and solve the resultant equation.

We now check our potential solution.

\begin{align*} \sqrt[3]{\highlight{10}+17}-\sqrt[3]{4 \cdot \highlight{10}-13}\amp=0\text{ ?}\\ \sqrt[3]{27}-\sqrt[3]{27}\amp=0\text{ ?}\\ 3-3\amp=0\,\checkmark \end{align*}

The solution set is $\{10\}$

###### 8

$x+3\sqrt{x}-4=0$

Solution

We begin by isolating the square root expression. We then square both sides of the equation and solve the resultant equation.

We now check our potential solutions.

\begin{align*} \highlight{16}+3\sqrt{\highlight{16}}-4\amp=0\text{ ?}\\ 16+3 \cdot 4-4\amp=0\text{ ?}\\ 24\amp=0\text{ not true!} \end{align*}
\begin{align*} \highlight{1}+3\sqrt{\highlight{1}}-4\amp=0\text{ ?}\\ 1+3 \cdot 1-4\amp=0\,\checkmark \end{align*}

The solution set is $\{1\}\text{.}$

###### 9

$2x+\sqrt{x}-3=0$

Solution

We begin by isolating the square root expression. We then square both sides of the equation and solve the resultant equation.

\begin{align*} 2x+\sqrt{x}-3\amp=0\\ 2x+\sqrt{x}-3\subtractright{2x}\addright{3}\amp=0\subtractright{2x}\addright{3}\\ \sqrt{x}\amp=-2x+3\\ \highlight{(}\sqrt{x}\highlight{)^2}\amp=\highlight{(}-2x+3\highlight{)^2}\\ x\amp=4x^2-12x+9\\ x\subtractright{x}\amp=4x^2-12x+9\subtractright{x}\\ 0\amp=4x^2-13x+9\\ 0\amp=4x^2-4x-9x+9\\ 0\amp=4x\highlightr{(x-1)}-9\highlightr{(x-1)}\\ 0\amp=(4x-9)\highlightr{(x-1)} \end{align*}
\begin{align*} 4x-9\amp=0 \amp\amp\text{ or }\amp x-1\amp=0\\ 4x-9\addright{9}\amp=0\addright{9} \amp\amp\text{ or }\amp x-1\addright{1}\amp=0\addright{1}\\ 4x\amp=9 \amp\amp\text{ or }\amp x\amp=1\\ \divideunder{4x}{4}\amp=\divideunder{9}{4} \amp\amp\text{ or }\amp x\amp=1\\ x\amp=\frac{9}{4} \amp\amp\text{ or }\amp x\amp=1 \end{align*}

We now check our potential solutions.

\begin{align*} 2 \cdot \highlight{\frac{9}{4}}+\sqrt{\highlight{\frac{9}{4}}}-3\amp=0\text{ ?}\\ \frac{9}{2}+\frac{3}{2}-3\amp=0\text{ ?}\\ 6-3\amp=0\text{ ?}\\ 3\amp=0\text{ untrue} \end{align*}
\begin{align*} 2 \cdot \highlight{1}+\sqrt{\highlight{1}}-3\amp=0\text{ ?}\\ 2+1-3\amp=0\,\checkmark \end{align*}

The solution set is $\{1\}\text{.}$

###### 10

$\sqrt{1+\sqrt{9-x}}+7=9$

Solution

We begin by isolating the larger square root expression and squaring both sides of the equation. We then repeat the process for the smaller square root expression and solve the resultant equation.

\begin{align*} \sqrt{1+\sqrt{9-x}}+7\amp=9\\ \sqrt{1+\sqrt{9-x}}+7\subtractright{7}\amp=9\subtractright{7}\\ \sqrt{1+\sqrt{9-x}}\amp=2\\ \highlight{(}\sqrt{1+\sqrt{9-x}}\highlight{)^2}\amp=\highlight{(}2\highlight{)^2}\\ 1+\sqrt{9-x}\amp=4\\ 1+\sqrt{9-x}\subtractright{1}\amp=4\subtractright{1}\\ \sqrt{9-x}\amp=3\\ \highlight{(}\sqrt{9-x}\highlight{)^2}\amp=\highlight{(}3\highlight{)^2}\\ 9-x\amp=9\\ 9-x\subtractright{9}\amp=9\subtractright{9}\\ -x\amp=0\\ \multiplyleft{-1}-x\amp=\multiplyleft{-1}0\\ x\amp=0 \end{align*}

Let's go ahead and check the value of $0$ in the original equation.

\begin{align*} \sqrt{1+\sqrt{9-\highlight{0}}}+7\amp=9\text{ ?}\\ \sqrt{1+\sqrt{9}}+7\amp=9\text{ ?}\\ \sqrt{4}+7\amp=9\text{ ?}\\ \sqrt{1+3}+7\amp=9\text{ ?}\\ 2+7\amp=9\,\checkmark \end{align*}

The solution set is $\{0\}\text{.}$

###### 11

$2\sqrt{y-8}-\sqrt{2y+12}=0$

Solution

We begin by moving the square root expressions to opposite sides of the equal sign and then squaring both sides of the equation.

We now check.

\begin{align*} 2\sqrt{\highlight{22}-8}-\sqrt{2(\highlight{22})+12}\amp=0\text{ ?}\\ 2\sqrt{14}-\sqrt{56}\amp=0\text{ ?}\\ 2\sqrt{14}-\sqrt{4 \cdot 14}\amp=0\text{ ?}\\ 2\sqrt{14}-2\sqrt{14}\amp=0\,\checkmark \end{align*}

The solution set is $\{22\}\text{.}$

###### 12

$1-\sqrt{x}=\sqrt{x-5}$

Solution

Since one of the square root expressions is already isolated, we can go ahead and square both sides of the equation.

\begin{align*} 1-\sqrt{x}\amp=\sqrt{x-5}\\ \highlight{(}1-\sqrt{x}\highlight{)^2}\amp=\highlight{(}\sqrt{x-5}\highlight{)^2}\\ 1-2\sqrt{x}+x\amp=x-5\\ 1-2\sqrt{x}+x\subtractright{1}\subtractright{x}\amp=x-5\subtractright{1}\subtractright{x}\\ -2\sqrt{x}\amp=-6\\ \multiplyleft{-\frac{1}{2}}-2\sqrt{x}\amp=\multiplyleft{-\frac{1}{2}}-6\\ \sqrt{x}\amp=3\\ \highlight{(}\sqrt{x}\highlight{)^2}\amp=\highlight{(}3\highlight{)^2}\\ x\amp=9 \end{align*}

We now check.

\begin{align*} 1-\sqrt{\highlight{9}}\amp=\sqrt{\highlight{9}-5}\text{ ?}\\ 1-3\amp=2\text{ ?}\\ -2\amp=2\text{ nope!} \end{align*}

The solution set is $\emptyset\text{.}$

###### 13

$2\sqrt{3x+6}-\sqrt{4x+9}=5$

Solution

We begin by isolating the expression $2\sqrt{3x+6}$ and the squaring both sides of the equation.

\begin{align*} 2\sqrt{3x+6}-\sqrt{4x+9}\amp=5\\ 2\sqrt{3x+6}-\sqrt{4x+9}\addright{\sqrt{4x+9}}\amp=5\addright{\sqrt{4x+9}}\\ 2\sqrt{3x+6}\amp=5+\sqrt{4x+9}\\ \highlight{(}2\sqrt{3x+6}\highlight{)^2}\amp=\highlight{(}5+\sqrt{4x+9}\highlight{)^2}\\ 4(3x+6)\amp=25+10\sqrt{4x+9}+4x+9\\ 12x+24\amp=10\sqrt{4x+9}+4x+34\\ 12x+24\subtractright{4x}\subtractright{34}\amp=25+10\sqrt{4x+9}+4x+9\subtractright{4x}\subtractright{34}\\ 8x-10\amp=10\sqrt{4x+9}\\ \multiplyleft{\frac{1}{2}}(8x-10)\amp=\multiplyleft{\frac{1}{2}}10\sqrt{4x+9}\\ 4x-5\amp=5\sqrt{4x+9}\\ \highlight{(}4x-5\highlight{)^2}\amp=\highlight{(}5\sqrt{4x+9}\highlight{)^2}\\ 16x^2-40x+25\amp=25(4x+9)\\ 16x^2-40x+25\amp=100x+225\\ 16x^2-40x+25\subtractright{100x}\subtractright{225}\amp=100x+225\subtractright{100x}\subtractright{225}\\ 16x^2-140x-200\amp=0\\ \multiplyleft{\frac{1}{4}}(16x^2-140x-200)\amp=\multiplyleft{\frac{1}{4}}0\\ 4x^2-35x-50\amp=0\\ 4x^2-40x+5x-50\amp=0\\ 4x(x-10)+5(x-10)\amp=0\\ (4x+5)(x-10)\amp=0 \end{align*}
\begin{align*} 4x+5\amp=0 \amp\amp\text{ or }\amp x-10\amp=0\\ 4x+5\subtractright{5}\amp=0\subtractright{5} \amp\amp\text{ or }\amp x-10\addright{10}\amp=0\addright{10}\\ 4x\amp=-5 \amp\amp\text{ or }\amp x\amp=10\\ \divideunder{4x}{4}\amp=\divideunder{-5}{4} \amp\amp\text{ or }\amp x\amp=10\\ x\amp=-\frac{5}{4} \amp\amp\text{ or }\amp x\amp=10 \end{align*}

Let's now check $-\frac{5}{4}\text{.}$

\begin{align*} 2\sqrt{3 \cdot \highlight{-\frac{5}{4}}+6}-\sqrt{4 \cdot \highlight{-\frac{5}{4}}+9}\amp=5\text{ ?}\\ 2\sqrt{\frac{9}{4}}-\sqrt{4}\amp=5\text{ ?}\\ 2 \cdot \frac{3}{2}-2\amp=5\text{ ?}\\ 3-2\amp=5\text{ :(} \end{align*}

Seriously? Hmm ... let's check $10\text{.}$

\begin{align*} 2\sqrt{3 \cdot \highlight{10}+6}-\sqrt{4 \cdot \highlight{10}+9}\amp=5\text{ ?}\\ 2\sqrt{36}-\sqrt{49}\amp=5\text{ ?}\\ 2 \cdot 6-7\amp=5\text{ ?}\\ 12-7\amp=5\,\checkmark \end{align*}

The solution set is $\{10\}\text{.}$

###### 14

$\sqrt{24-x}+8=\sqrt{9-x}+11$

Solution

We begin by isolating $\sqrt{24-x}$ on the left side of the equal sign and then squaring both sides of the equation.

\begin{align*} \sqrt{24-x}+8\amp=\sqrt{9-x}+11\\ \sqrt{24-x}+8\subtractright{8}\amp=\sqrt{9-x}+11\subtractright{8}\\ \sqrt{24-x}\amp=\sqrt{9-x}+3\\ \highlight{(}\sqrt{24-x}\highlight{)^2}\amp=\highlight{(}\sqrt{9-x}+3\highlight{)^2}\\ 24-x\amp=9-x+6\sqrt{9-x}+9\\ 24-x\amp=6\sqrt{9-x}+18-x\\ 24-x\amp\subtractright{18}\addright{x}=6\sqrt{9-x}+18-x\subtractright{18}\addright{x}\\ 6\amp=6\sqrt{9-x}\\ \divideunder{6}{6}\amp=\divideunder{6\sqrt{9-x}}{6}\\ 1\amp=\sqrt{9-x}\\ \highlight{(}1\highlight{)^2}\amp=\highlight{(}\sqrt{9-x}\highlight{)^2}\\ 1\amp=9-x\\ 1\subtractright{9}\amp=9-x\subtractright{9}\\ -8\amp=-x\\ \multiplyleft{-1}-8\amp=\multiplyleft{-1}-x\\ 8\amp=x \end{align*}

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