## Section14.5Radical Functions and Equations Chapter Review

In Section 14.1 we covered the square root and other root functions. We learned how to find the domain of radical functions algebraically and the range graphically. We also saw the distance formula which is an application of square roots.

###### Example14.5.1The Square Root Function

Algebraically find the domain of the function $f$ where $f(x)=\sqrt{3x-1}+2$ and then find the range by making a graph.

Explanation

Using Fact 14.1.29 to find the function's domain, we set the radicand greater than or equal to zero and solve:

\begin{align*} 3x-1\amp\ge0\\ 3x\amp\ge1\\ x\amp\ge\frac{1}{3} \end{align*}

The function's domain is $\left[\frac{1}{3},\infty\right)$ in interval notation.

To find the function's range, we use technology to look at a graph of the function. The graph shows that the function's range is $[2,\infty)\text{.}$ The graph also verifies the function's domain is indeed $\left[\frac{1}{3},\infty\right)\text{.}$

###### Example14.5.3The Distance Formula

Find the distance between $(6,-13)$ and $(-4,17)\text{.}$

Explanation

To calculate the distance between $(6,-13)$ and $(-4,17)\text{,}$ we use the distance formula. It's good practice to mark each value with the corresponding variables in the formula:

\begin{equation*} (\stackrel{x_1}{6},\stackrel{y_1}{-13}), (\stackrel{x_2}{-4},\stackrel{y_2}{17}) \end{equation*}

We have:

\begin{align*} d\amp=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ d\amp=\sqrt{\left(\substitute{-4}-\substitute{6}\right)^2+\left(\substitute{17}-\substitute{(-13)}\right)^2}\\ d\amp=\sqrt{\left(-10\right)^2+\left(30\right)^2}\\ d\amp=\sqrt{100+900}\\ d\amp=\sqrt{1000}\\ d\amp=\sqrt{100}\cdot\sqrt{10}\\ d\amp=10\sqrt{10} \end{align*}

The distance between $(6,-13)$ and $(-4,17)$ is $10\sqrt{10}$ or approximately $31.62$ units.

###### Example14.5.4The Cube Root Function

Algebraically find the domain and graphically find the range of the function $g$ where $g(x)=-2\sqrt[3]{x+6}-1\text{.}$

Explanation

First note that the index of the function $g$ for $g(x)=-2\sqrt[3]{x+6}-1$ is odd. By Fact 14.1.29, the domain is $(-\infty,\infty)\text{.}$

To find the function's range, we use technology to graph the function. According to the graph, the function's range is also $(-\infty,\infty)\text{.}$

###### Example14.5.6Other Roots

Algebraically find the domain and graphically find the range of the function $h$ where $h(x)=8-\frac{5}{2}\sqrt[4]{6-2x}\text{.}$

Explanation

First note that the index of this function is $4\text{,}$ which is even. By Fact 14.1.29, to find the domain of this function we must set the radicand greater or equal to zero and solve.

\begin{align*} 6-2x\amp\ge0\\ -2x\amp\ge\subtractright{6}\\ x\amp\mathbin{\highlight{\le}}\divideunder{-6}{-2}\\ x\amp\le3 \end{align*}

So, the domain of $h$ is $(-\infty,3]\text{.}$

To find the function's range, we use technology to graph the function. By the graph, the function's range is $(-\infty,8]\text{.}$

### Subsection14.5.2Radical Expressions and Rational Exponents

In Section 14.2 we learned the rational exponent rule and added it to our list of exponent rules.

###### Example14.5.8Radical Expressions and Rational Exponents

Simplify the expressions using Fact 14.2.2 or Fact 14.2.10.

1. $100^{\sfrac{1}{2}}$

2. $(-64)^{-\sfrac{1}{3}}$

3. $-81^{\sfrac{3}{4}}$

4. $\left(-\frac{1}{27}\right)^{\sfrac{2}{3}}$

Explanation
1. \begin{aligned}[t] 100^{\sfrac{1}{2}}\amp=\left(\sqrt{100}\right)\\ \amp=10 \end{aligned}

2. \begin{aligned}[t] (-64)^{-\sfrac{1}{3}}\amp=\frac{1}{(-64)^{\sfrac{1}{3}}}\\ \amp=\frac{1}{\left(\sqrt[3]{(-64)}\right)}\\ \amp=\frac{1}{-4} \end{aligned}

3. \begin{aligned}[t] -81^{\sfrac{3}{4}}\amp=-\left(\sqrt[4]{81}\right)^3\\ \amp=-3^3\\ \amp=-27 \end{aligned}

4. In this problem the negative can be associated with either the numerator or the denominator, but not both. We choose the numerator.

\begin{align*} \left(-\frac{1}{27}\right)^{\sfrac{2}{3}}\amp=\left(\sqrt[3]{-\frac{1}{27}}\right)^2\\ \amp=\left(\frac{\sqrt[3]{-1}}{\sqrt[3]{27}}\right)^2\\ \amp=\left(\frac{-1}{3}\right)^2\\ \amp=\frac{(-1)^2}{(3)^2}\\ \amp=\frac{1}{9} \end{align*}
###### Example14.5.9More Expressions with Rational Exponents

Use exponent properties in List 14.2.15 to simplify the expressions, and write all final versions using radicals.

1. $7z^{\sfrac{5}{9}}$

2. $\frac{5}{4}x^{-\sfrac{2}{3}}$

3. $\left(-9q^5\right)^{\sfrac{4}{5}}$

4. $\sqrt{y^5}\cdot\sqrt[4]{y^2}$

5. $\frac{\sqrt{t^3}}{\sqrt[3]{t^2}}$

6. $\sqrt{\sqrt[3]{x}}$

7. $5\left(4+a^{\sfrac{1}{2}}\right)^2$

8. $-6\left(2p^{-\sfrac{5}{2}}\right)^{\sfrac{3}{5}}$

Explanation
1. \begin{aligned}[t] 7z^{\sfrac{5}{9}}\amp=7\sqrt[9]{z^5} \end{aligned}

2. \begin{aligned}[t] \frac{5}{4}x^{-\sfrac{2}{3}}\amp=\frac{5}{4}\cdot\frac{1}{x^{\sfrac{2}{3}}}\\ \amp=\frac{5}{4}\cdot\frac{1}{\sqrt[3]{x^2}}\\ \amp=\frac{5}{4\sqrt[3]{x^2}} \end{aligned}

3. \begin{aligned}[t] \left(-9q^5\right)^{\sfrac{4}{5}}\amp=\left(-9\right)^{\sfrac{4}{5}}\cdot\left(q^5\right)^{\sfrac{4}{5}}\\ \amp=\left(-9\right)^{\sfrac{4}{5}}\cdot q^{5\cdot\sfrac{4}{5}}\\ \amp=\left(\sqrt[5]{-9}\right)^4\cdot q^{4}\\ \amp=\left(q\sqrt[5]{-9}\right)^4 \end{aligned}

4. \begin{aligned}[t] \sqrt{y^5}\cdot\sqrt[4]{y^2}\amp=y^{\sfrac{5}{2}}\cdot y^{\sfrac{2}{4}}\\ \amp=y^{\sfrac{5}{2}+\sfrac{2}{4}}\\ \amp=y^{\sfrac{10}{4}+\sfrac{1}{4}}\\ \amp=x^{\sfrac{11}{4}}\\ \amp=\sqrt[4]{x^{11}} \end{aligned}

5. \begin{aligned}[t] \frac{\sqrt{t^3}}{\sqrt[3]{t^2}}\amp=\frac{t^{\sfrac{3}{2}}}{t^{\sfrac{2}{3}}}\\ \amp=t^{\sfrac{3}{2}-\sfrac{2}{3}}\\ \amp=t^{\sfrac{9}{6}-\sfrac{4}{6}}\\ \amp=t^{\sfrac{5}{6}}\\ \amp=\sqrt[6]{t^5} \end{aligned}

6. \begin{aligned}[t] \sqrt{\sqrt[3]{x}}\amp=\sqrt{x^{\sfrac{1}{3}}}\\ \amp=\left(x^{\sfrac{1}{3}}\right)^{\sfrac{1}{2}}\\ \amp=x^{\sfrac{1}{3}\cdot\sfrac{1}{2}}\\ \amp=x^{\sfrac{1}{6}}\\ \amp=\sqrt[6]{x} \end{aligned}

7. \begin{aligned}[t] 5\left(4+a^{\sfrac{1}{2}}\right)^2\amp=5\left(4+a^{\sfrac{1}{2}}\right)\left(4+a^{\sfrac{1}{2}}\right)\\ \amp=5\left(4^2+2\cdot4\cdot a^{\sfrac{1}{2}}+\left(a^{\sfrac{1}{2}}\right)^2\right)\\ \amp=5\left(16+8a^{\sfrac{1}{2}}+a^{\sfrac{1}{2}\cdot 2}\right)\\ \amp=5\left(16+8a^{\sfrac{1}{2}}+a\right)\\ \amp=5\left(16+8\sqrt{a}+a\right)\\ \amp=80+40\sqrt{a}+5a \end{aligned}

8. \begin{aligned}[t] -6\left(2p^{-\sfrac{5}{2}}\right)^{\sfrac{3}{5}}\amp=-6\cdot2^{\sfrac{3}{5}}\cdot p^{-\sfrac{5}{2}\cdot\sfrac{3}{5}}\\ \amp=-6\cdot2^{\sfrac{3}{5}}\cdot p^{-\sfrac{3}{2}}\\ \amp=-\frac{6\cdot 2^{\sfrac{3}{5}}}{p^{\sfrac{3}{2}}}\\ \amp=-\frac{6\sqrt[5]{2^3}}{\sqrt{p^3}}\\ \amp=-\frac{6\sqrt[5]{8}}{\sqrt{p^3}} \end{aligned}

### Subsection14.5.3More on Rationalizing the Denominator

In Section 14.3 we covered how to rationalize the denominator when it contains a single square root or a binomial.

###### Example14.5.10A Review of Rationalizing the Denominator

Rationalize the denominator of the expressions.

1. $\frac{12}{\sqrt{3}}$

2. $\frac{\sqrt{5}}{\sqrt{75}}$

Explanation
1. \begin{align*} \frac{12}{\sqrt{3}}\amp=\frac{12}{\sqrt{3}}\multiplyright{\frac{\sqrt{3}}{\sqrt{3}}}\\ \amp=\frac{12\sqrt{3}}{3}\\ \amp=4\sqrt{3} \end{align*}
2. First we will simplify $\sqrt{75}\text{.}$

\begin{align*} \frac{\sqrt{5}}{\sqrt{75}}\amp=\frac{\sqrt{5}}{\sqrt{25\cdot 3}}\\ \amp=\frac{\sqrt{5}}{\sqrt{25}\cdot\sqrt{3}}\\ \amp=\frac{\sqrt{5}}{5\sqrt{3}}\\ \end{align*}

Now we can rationalize the denominator by multiplying the numerator and denominator by $\sqrt{3}\text{.}$

\begin{align*} \amp=\frac{\sqrt{5}}{5\sqrt{3}}\multiplyright{\frac{\sqrt{3}}{\sqrt{3}}}\\ \amp=\frac{\sqrt{15}}{5\cdot 3}\\ \amp=\frac{\sqrt{15}}{15} \end{align*}
###### Example14.5.11Rationalize Denominator with Difference of Squares Formula

Rationalize the denominator in $\frac{\sqrt{6}-\sqrt{5}}{\sqrt{3}+\sqrt{2}}\text{.}$

Explanation

To remove radicals in $\sqrt{3}+\sqrt{2}$ with the difference of squares formula, we multiply it with $\sqrt{3}-\sqrt{2}\text{.}$

\begin{align*} \frac{\sqrt{6}-\sqrt{5}}{\sqrt{3}+\sqrt{2}}\amp=\frac{\sqrt{6}-\sqrt{5}}{\sqrt{3}+\sqrt{2}}\multiplyright{\frac{\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)}}\\ \amp=\frac{\sqrt{6}\multiplyright{\sqrt{3}}-\sqrt{6}\multiplyright{\sqrt{2}}-\sqrt{5}\multiplyright{\sqrt{3}}-\sqrt{5}\multiplyright{-\sqrt{2}}}{\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2}\\ \amp=\frac{\sqrt{18}-\sqrt{12}-\sqrt{15}+\sqrt{10}}{9-4}\\ \amp=\frac{3\sqrt{2}-2\sqrt{3}-\sqrt{15}+\sqrt{10}}{5} \end{align*}

In Section 14.4 we covered solving equations that contain a radical. We learned about extraneous solutions and the need to check our solutions.

Solve for $r$ in $r=9+\sqrt{r+3}\text{.}$

Explanation

We will isolate the radical first, and then square both sides.

\begin{align*} r\amp=9+\sqrt{r+3}\\ r-9\amp=\sqrt{r+3}\\ \left(r-9\right)^{\highlight{2}}\amp=\left(\sqrt{r+3}\right)^{\highlight{2}}\\ r^2-18r+81\amp=r+3\\ r^2-19r+78\amp=0\\ (r-6)(r-13)\amp=0 \end{align*}
\begin{align*} r-6\amp=0\amp\amp\text{ or }r-13\amp=0\\ r\amp=6\amp\amp\text{ or }r\amp=13 \end{align*}

Because we squared both sides of an equation, we must check both solutions.

\begin{align*} \substitute{6}\amp\stackrel{?}{=}9+\sqrt{\substitute{6}+3}\amp\substitute{13}\amp\stackrel{?}{=}9+\sqrt{\substitute{13}+3}\\ 6\amp\stackrel{?}{=}9+\sqrt{9}\amp13\amp\stackrel{?}{=}9+\sqrt{16}\\ 6\amp\stackrel{\text{no}}{=}9+3\amp13\amp\stackrel{\checkmark}{=}9+4 \end{align*}

It turns out $6$ is an extraneous solution and $13$ is a valid solution. So the equation has one solution: $13\text{.}$ The solution set is $\{13\}\text{.}$

###### Example14.5.13Solving Radical Equations that Require Squaring Twice

Solve the equation $\sqrt{t+9}=-1-\sqrt{t}$ for $t\text{.}$

Explanation

We cannot isolate two radicals, so we will simply square both sides, and later try to isolate the remaining radical.

\begin{align*} \sqrt{t+9}\amp=-1-\sqrt{t}\\ \left(\sqrt{t+9}\right)^\highlight{2}\amp=\left(-1-\sqrt{t}\right)^\highlight{2}\\ t+9\amp=1+2\sqrt{t}+t \amp\text{ after expanding the binomial squared}\\ 9\amp=1+2\sqrt{t}\\ 8\amp=2\sqrt{t}\\ 4\amp=\sqrt{t}\\ (4)^\highlight{2}\amp=\left(\sqrt{t}\right)^\highlight{2}\\ 16\amp=t \end{align*}

Because we squared both sides of an equation, we must check the solution by substituting $\substitute{16}$ into $\sqrt{t+9}=-1-\sqrt{t}\text{,}$ and we have:

\begin{align*} \sqrt{t+9}\amp=-1-\sqrt{t}\\ \sqrt{\substitute{16}+9}\amp\stackrel{?}{=}-1-\sqrt{16}\\ \sqrt{25}\amp\stackrel{?}{=}-1-4\\ 5\amp\stackrel{\text{no}}{=}-5 \end{align*}

Our solution did not check so there is no solution to this equation. The solution set is the empty set, which can be denotes $\{\text{ }\}$ or $\emptyset\text{.}$

### Subsection14.5.5Exercises

###### 1

Find the domain of the function.

$H(x)={\sqrt{8-x}}$

###### 2

Find the domain of the function.

$H(x)={\sqrt{5-x}}$

###### 3

Find the domain of the function.

$K(x)=\sqrt[3]{{-9x+10}}$

###### 4

Find the domain of the function.

$f(x)=\sqrt[3]{{5x-2}}$

###### 5

Find the domain of the function.

$g(x)=\sqrt[4]{{18-3x}}$

###### 6

Find the domain of the function.

$h(x)=\sqrt[4]{{-12-2x}}$

###### 7

Use technology to find the range of the function $h$ defined by $h(x)={\sqrt{2-x}+1}\text{.}$

###### 8

Use technology to find the range of the function $F$ defined by $F(x)={\sqrt{-2-x}-5}\text{.}$

###### 9

If an object is dropped with no initial velocity, the time since the drop, in seconds, can be calculated by the function

\begin{equation*} T(h) = \sqrt{\frac{2h}{g}} \end{equation*}

where $h$ is the distance the object traveled in feet. The variable $g$ is the gravitational acceleration on earth, and we can round it to $32 \frac{ft}{s^2}$ for this problem.

1. After seconds since the release, the object would have traveled $35$ feet.

2. After $5$ seconds since the release, the object would have traveled feet.

###### 10

If an object is dropped with no initial velocity, the time since the drop, in seconds, can be calculated by the function

\begin{equation*} T(h) = \sqrt{\frac{2h}{g}} \end{equation*}

where $h$ is the distance the object traveled in feet. The variable $g$ is the gravitational acceleration on earth, and we can round it to $32 \frac{ft}{s^2}$ for this problem.

1. After seconds since the release, the object would have traveled $40$ feet.

2. After $3.6$ seconds since the release, the object would have traveled feet.

###### 11

Find the distance between the points $(-8,-2)$ and $(57,70)\text{.}$

###### 12

Find the distance between the points $(-10,-15)$ and $(45,33)\text{.}$

###### 13

Without using a calculator, evaluate the expression.

$\displaystyle{\sqrt[4]{3^{3}}=}$

###### 14

Without using a calculator, evaluate the expression.

$\displaystyle{\sqrt[3]{13^{2}}=}$

###### 15

Without using a calculator, evaluate the expression.

$\displaystyle{ \sqrt[3]{{-{\frac{8}{125}}}}= }$ .

###### 16

Without using a calculator, evaluate the expression.

$\displaystyle{ \sqrt[3]{{-{\frac{1}{8}}}}= }$ .

###### 17

Use rational exponents to write the expression.

$\displaystyle{\sqrt[5]{3 m + 9}=}$

###### 18

Use rational exponents to write the expression.

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

###### 19

Convert the expression to radical notation.

$\displaystyle{{14^{\frac{1}{5}}a^{\frac{4}{5}}}}$ =

###### 20

Convert the expression to radical notation.

$\displaystyle{{3^{\frac{1}{3}}b^{\frac{2}{3}}}}$ =

###### 21

Convert $c^{-\frac{5}{7}}$ to a radical.

• $-\sqrt[7]{c^{5}}$

• $\frac{1}{\sqrt[5]{c^{7}}}$

• $\frac{1}{\sqrt[7]{c^{5}}}$

• $-\sqrt[5]{c^{7}}$

###### 22

Convert $x^{-\frac{3}{8}}$ to a radical.

• $\frac{1}{\sqrt[3]{x^{8}}}$

• $-\sqrt[8]{x^{3}}$

• $\frac{1}{\sqrt[8]{x^{3}}}$

• $-\sqrt[3]{x^{8}}$

###### 23

$\displaystyle{\frac{\sqrt{25 y}}{\sqrt[10]{y^{3}}}=}$

###### 24

$\displaystyle{\frac{\sqrt{9 z}}{\sqrt[6]{z^{5}}}=}$

###### 25

$\displaystyle{\sqrt[5]{t}\cdot\sqrt[10]{t^{3}}=}$

###### 26

$\displaystyle{\sqrt[5]{m}\cdot\sqrt[10]{m^{3}}=}$

###### 27

$\displaystyle{\sqrt{a}\sqrt[7]{a}=}$

###### 28

$\displaystyle{\sqrt{t}\sqrt[8]{t}=}$

###### 29

Rationalize the denominator and simplify the expression.

$\displaystyle{ \frac{9}{{\sqrt{72}}} = }$

###### 30

Rationalize the denominator and simplify the expression.

$\displaystyle{ \frac{4}{{\sqrt{216}}} = }$

###### 31

Rationalize the denominator and simplify the expression.

$\displaystyle{ \sqrt{\frac{7}{12}} = }$

###### 32

Rationalize the denominator and simplify the expression.

$\displaystyle{ \sqrt{\frac{11}{28}} = }$

###### 33

Rationalize the denominator and simplify the expression.

$\displaystyle{\dfrac{4}{\sqrt{23}+6}=}$

###### 34

Rationalize the denominator and simplify the expression.

$\displaystyle{\dfrac{5}{\sqrt{7}+9}=}$

###### 35

Rationalize the denominator and simplify the expression.

$\displaystyle{\dfrac{\sqrt{5}-10}{\sqrt{7}+9}=}$

###### 36

Rationalize the denominator and simplify the expression.

$\displaystyle{\dfrac{\sqrt{3}-11}{\sqrt{13}+6}=}$

###### 37

Solve the equation.

$\displaystyle{ {r} = {\sqrt{r+2}+4} }$

###### 38

Solve the equation.

$\displaystyle{ {t} = {\sqrt{t+3}+3} }$

###### 39

Solve the equation.

$\displaystyle{ {\sqrt{t+8}} = {\sqrt{t}-4} }$

###### 40

Solve the equation.

$\displaystyle{ {\sqrt{x+8}} = {\sqrt{x}+2} }$

###### 41

Solve the equation.

$\displaystyle{ {\sqrt{x}+110} = {x} }$

###### 42

Solve the equation.

$\displaystyle{ {\sqrt{y}+56} = {y} }$

###### 43

Solve the equation.

$\displaystyle{ {y} = {\sqrt{y+3}+17} }$

###### 44

Solve the equation.

$\displaystyle{ {r} = {\sqrt{r+1}+109} }$

###### 45

Solve the equation.

$\displaystyle{ {\sqrt{66-r}} = {r+6} }$

###### 46

Solve the equation.

$\displaystyle{ {\sqrt{29-r}} = {r+1} }$

###### 47

Use technology to solve the equation ${\sqrt{x-1.9}}={\sqrt{x}-0.2}\text{.}$

###### 48

Use technology to solve the equation ${\sqrt{x+2.3}}={\sqrt{x}-3}\text{.}$

###### 49

According to the Pythagorean Theorem, the length $c$ of the hypothenuse of a rectangular triangle can be found through the following equation:

\begin{equation*} {c} = {\sqrt{a^{2}+b^{2}}} \end{equation*}

Solve the equation for the length $a$ of one of the triangle’s legs.

$a =$ .

###### 50

In a coordinate system, the distance $r$ from a point $(x,y)$ to the origin $(0,0)$ is given by the following equation:

\begin{equation*} {r} = {\sqrt{x^{2}+y^{2}}} \end{equation*}

Solve the equation for the coordinate $y\text{.}$ Assume that $y$ is positive.

$y =$ .

###### 51

According to the Pythagorean Theorem, the length $c$ of the hypothenuse of a rectangular triangle can be found through the following equation.

\begin{equation*} {c} = {\sqrt{a^{2}+b^{2}}} \end{equation*}

If a rectangular triangle has a hypothenuse of ${13\ {\rm ft}}$ and one leg is ${12\ {\rm ft}}$ long, how long is the third side of the triangle?

The third side of the triangle is long.

###### 52

According to the Pythagorean Theorem, the length $c$ of the hypothenuse of a rectangular triangle can be found through the following equation.

\begin{equation*} {c} = {\sqrt{a^{2}+b^{2}}} \end{equation*}

If a rectangular triangle has a hypothenuse of ${17\ {\rm ft}}$ and one leg is ${15\ {\rm ft}}$ long, how long is the third side of the triangle?

The third side of the triangle is long.

###### 53

A pendulum has the length $L$ ft. The time period $T$ that it takes to once swing back and forth is $6$ s. Use the following formula to find its length.

\begin{equation*} {T} = {2\pi \sqrt{\frac{L}{32}}} \end{equation*}

The pendulum is long.

###### 54

A pendulum has the length $L$ ft. The time period $T$ that it takes to once swing back and forth is $8$ s. Use the following formula to find its length.

\begin{equation*} {T} = {2\pi \sqrt{\frac{L}{32}}} \end{equation*}

The pendulum is long.