ORCCA Radical Expressions and Equations Chapter Review

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Section 6.5 Radical Expressions and Equations Chapter Review

Subsection 6.5.1 Square and $n$ th Root Properties

In Section 6.1 we defined the square root $\sqrt{x}$ and $n$ th root $\sqrt[n]{x}$ radicals. When $x$ is positive, the expression $\sqrt[n]{x}$ means a positive number $r\text{,}$ where $\overbrace{r\cdot r\cdot\cdots\cdot r}^{n\text{ times}}=x\text{.}$ The square root $\sqrt{x}$ is just the case where $n=2\text{.}$

When $x$ is negative, $\sqrt[n]{x}$ might not be defined. It depends on whether or not $n$ is an even number. When $x$ is negative and $n$ is odd, $\sqrt[n]{x}$ is a negative number where $\overbrace{r\cdot r\cdot\cdots\cdot r}^{n\text{ times}}=x\text{.}$

permalinkThere are two helpful rules for simplifying radicals.

List 6.5.1. Rules of Radicals for Multiplication and Division

If $a$ and $b$ are positive real numbers, and $m$ is a positive integer , then we have the following rules:

Root of a Product Rule: $\sqrt[m]{a\cdot b} = \sqrt[m]{a} \cdot \sqrt[m]{b}$
Root of a Quotient Rule: $\sqrt[m]{\frac{a}{b}} = \frac{\sqrt[m]{a}}{\sqrt[m]{b}}$ as long as $b \neq 0$

Checkpoint 6.5.2.

Subsection 6.5.2 Rationalizing the Denominator

permalinkIn Section 6.2 we covered how to rationalize the denominator when it contains a single square root or a binomial with a square root term.

Example 6.5.3.

Rationalize the denominator of the expressions.

$\frac{12}{\sqrt{3}}$
$\frac{\sqrt{5}}{\sqrt{75}}$

\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*}
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*}

Example 6.5.4. Rationalize Denominator Using the Difference of Squares Formula.

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

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}}{3-2}\\ \amp=\frac{3\sqrt{2}-2\sqrt{3}-\sqrt{15}+\sqrt{10}}{1}\\ \amp=3\sqrt{2}-2\sqrt{3}-\sqrt{15}+\sqrt{10} \end{align*}

Subsection 6.5.3 Radical Expressions and Rational Exponents

permalinkIn Section 6.3 we learned the rational exponent rule and added it to our list of exponent rules.

Example 6.5.5. Radical Expressions and Rational Exponents.

Simplify the expressions using Fact 6.3.2 or Fact 6.3.9.

$100^{\sfrac{1}{2}}$
$(-64)^{-\sfrac{1}{3}}$
$-81^{\sfrac{3}{4}}$
$\left(-\frac{1}{27}\right)^{\sfrac{2}{3}}$

\(\begin{aligned}[t] 100^{\sfrac{1}{2}}\amp=\left(\sqrt{100}\right)\\ \amp=10 \end{aligned}\)
\(\begin{aligned}[t] (-64)^{-\sfrac{1}{3}}\amp=\frac{1}{(-64)^{\sfrac{1}{3}}}\\ \amp=\frac{1}{\left(\sqrt[3]{(-64)}\right)}\\ \amp=\frac{1}{-4} \end{aligned}\)
\(\begin{aligned}[t] -81^{\sfrac{3}{4}}\amp=-\left(\sqrt[4]{81}\right)^3\\ \amp=-3^3\\ \amp=-27 \end{aligned}\)
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*}

Example 6.5.6. More Expressions with Rational Exponents.

Use exponent properties in List 6.3.14 to simplify the expressions, and write all final versions using radicals.

$7z^{\sfrac{5}{9}}$
$\frac{5}{4}x^{-\sfrac{2}{3}}$
$\left(-9q^5\right)^{\sfrac{4}{5}}$
$\sqrt{y^5}\cdot\sqrt[4]{y^2}$
$\frac{\sqrt{t^3}}{\sqrt[3]{t^2}}$
$\sqrt{\sqrt[3]{x}}$
$5\left(4+a^{\sfrac{1}{2}}\right)^2$
$-6\left(2p^{-\sfrac{5}{2}}\right)^{\sfrac{3}{5}}$

\(\begin{aligned}[t] 7z^{\sfrac{5}{9}}\amp=7\sqrt[9]{z^5} \end{aligned}\)
\(\begin{aligned}[t] \frac{5}{4}x^{-\sfrac{2}{3}}\amp=\frac{5}{4}\cdot\frac{1}{x^{\sfrac{2}{3}}}\\ \amp=\frac{5}{4}\cdot\frac{1}{\sqrt[3]{x^2}}\\ \amp=\frac{5}{4\sqrt[3]{x^2}} \end{aligned}\)
\(\begin{aligned}[t] \left(-9q^5\right)^{\sfrac{4}{5}}\amp=\left(-9\right)^{\sfrac{4}{5}}\cdot\left(q^5\right)^{\sfrac{4}{5}}\\ \amp=\left(-9\right)^{\sfrac{4}{5}}\cdot q^{5\cdot\sfrac{4}{5}}\\ \amp=\left(\sqrt[5]{-9}\right)^4\cdot q^{4}\\ \amp=\left(q\sqrt[5]{-9}\right)^4 \end{aligned}\)
\(\begin{aligned}[t] \sqrt{y^5}\cdot\sqrt[4]{y^2}\amp=y^{\sfrac{5}{2}}\cdot y^{\sfrac{2}{4}}\\ \amp=y^{\sfrac{5}{2}+\sfrac{2}{4}}\\ \amp=y^{\sfrac{10}{4}+\sfrac{2}{4}}\\ \amp=y^{\sfrac{12}{4}}\\ \amp=y^3 \end{aligned}\)
\(\begin{aligned}[t] \frac{\sqrt{t^3}}{\sqrt[3]{t^2}}\amp=\frac{t^{\sfrac{3}{2}}}{t^{\sfrac{2}{3}}}\\ \amp=t^{\sfrac{3}{2}-\sfrac{2}{3}}\\ \amp=t^{\sfrac{9}{6}-\sfrac{4}{6}}\\ \amp=t^{\sfrac{5}{6}}\\ \amp=\sqrt[6]{t^5} \end{aligned}\)
\(\begin{aligned}[t] \sqrt{\sqrt[3]{x}}\amp=\sqrt{x^{\sfrac{1}{3}}}\\ \amp=\left(x^{\sfrac{1}{3}}\right)^{\sfrac{1}{2}}\\ \amp=x^{\sfrac{1}{3}\cdot\sfrac{1}{2}}\\ \amp=x^{\sfrac{1}{6}}\\ \amp=\sqrt[6]{x} \end{aligned}\)
\(\begin{aligned}[t] 5\left(4+a^{\sfrac{1}{2}}\right)^2\amp=5\left(4+a^{\sfrac{1}{2}}\right)\left(4+a^{\sfrac{1}{2}}\right)\\ \amp=5\left(4^2+2\cdot4\cdot a^{\sfrac{1}{2}}+\left(a^{\sfrac{1}{2}}\right)^2\right)\\ \amp=5\left(16+8a^{\sfrac{1}{2}}+a^{\sfrac{1}{2}\cdot 2}\right)\\ \amp=5\left(16+8a^{\sfrac{1}{2}}+a\right)\\ \amp=5\left(16+8\sqrt{a}+a\right)\\ \amp=80+40\sqrt{a}+5a \end{aligned}\)
\(\begin{aligned}[t] -6\left(2p^{-\sfrac{5}{2}}\right)^{\sfrac{3}{5}}\amp=-6\cdot2^{\sfrac{3}{5}}\cdot p^{-\sfrac{5}{2}\cdot\sfrac{3}{5}}\\ \amp=-6\cdot2^{\sfrac{3}{5}}\cdot p^{-\sfrac{3}{2}}\\ \amp=-\frac{6\cdot 2^{\sfrac{3}{5}}}{p^{\sfrac{3}{2}}}\\ \amp=-\frac{6\sqrt[5]{2^3}}{\sqrt{p^3}}\\ \amp=-\frac{6\sqrt[5]{8}}{\sqrt{p^3}} \end{aligned}\)

Subsection 6.5.4 Solving Radical Equations

permalinkIn Section 6.4 we covered solving equations that contain a radical. We learned about extraneous solutions and the need to check our solutions.

Example 6.5.7. Solving Radical Equations.

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

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 \(\{13\}\text{.}\)

Example 6.5.8. Solving Radical Equations that Require Squaring Twice.

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

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 denoted \(\{\text{ }\}\) or \(\emptyset\text{.}\)

Exercises 6.5.5 Exercises

Square Root and $n$ th Root

1.

Evaluate the following.

$\displaystyle{\sqrt{{{\frac{1}{100}}}}={}}$ .

2.

Evaluate the following.

$\displaystyle{\sqrt{{{\frac{4}{121}}}}={}}$ .

3.

Evaluate the following.

$-\sqrt{16}={}$ .

4.

Evaluate the following.

$-\sqrt{25}={}$ .

5.

Simplify the radical expression or state that it is not a real number.

$\displaystyle{ \frac{{\sqrt{48}}}{{\sqrt{3}}} =}$

6.

Simplify the radical expression or state that it is not a real number.

$\displaystyle{ \frac{{\sqrt{32}}}{{\sqrt{2}}} =}$

7.

Simplify the radical expression or state that it is not a real number.

$\displaystyle{ {\sqrt{250}} = }$

8.

Simplify the radical expression or state that it is not a real number.

$\displaystyle{ {\sqrt{99}} = }$

9.

Simplify the expression.

$9\sqrt{13} \cdot 9\sqrt{{121}} =$

10.

Simplify the expression.

$9\sqrt{3} \cdot 7\sqrt{{4}} =$

11.

Simplify the expression.

$\displaystyle{ \sqrt{\frac{5}{2}} \cdot \sqrt{\frac{7}{2}} =}$

12.

Simplify the expression.

$\displaystyle{ \sqrt{\frac{7}{3}} \cdot \sqrt{\frac{1}{3}} =}$

13.

Simplify the expression.

$\displaystyle{{13\sqrt{10}} - {14\sqrt{10}} =}$

14.

Simplify the expression.

$\displaystyle{{14\sqrt{5}} - {15\sqrt{5}} =}$

15.

Simplify the expression.

$\displaystyle{{\sqrt{180}} + {\sqrt{45}} =}$

16.

Simplify the expression.

$\displaystyle{{\sqrt{80}} + {\sqrt{125}} =}$

17.

Simplify ${\sqrt[6]{64}}\text{.}$

18.

Simplify ${\sqrt[3]{64}}\text{.}$

19.

Simplify ${\sqrt[3]{-8}}\text{.}$

20.

Simplify ${\sqrt[3]{-8}}\text{.}$

21.

Simplify ${\sqrt[4]{-16}}\text{.}$

22.

Simplify ${\sqrt[4]{-81}}\text{.}$

23.

Simplify ${\sqrt[4]{144}}\text{.}$

24.

Simplify ${\sqrt[3]{135}}\text{.}$

25.

Simplify ${\sqrt[3]{\frac{11}{8}}}\text{.}$

26.

Simplify ${\sqrt[6]{\frac{9}{64}}}\text{.}$

27.

Simplify ${\sqrt[3]{\frac{40}{27}}}\text{.}$

28.

Simplify ${\sqrt[3]{\frac{56}{125}}}\text{.}$

Rationalizing the Denominator

29.

Rationalize the denominator and simplify the expression.

$\displaystyle{ \frac{2}{{\sqrt{252}}} = }$

30.

Rationalize the denominator and simplify the expression.

$\displaystyle{ \frac{6}{{\sqrt{112}}} = }$

31.

Rationalize the denominator and simplify the expression.

$\displaystyle{ \sqrt{\frac{2}{27}} = }$

32.

Rationalize the denominator and simplify the expression.

$\displaystyle{ \sqrt{\frac{5}{112}} = }$

33.

Rationalize the denominator and simplify the expression.

$\displaystyle{\dfrac{6}{\sqrt{15}+8}=}$

34.

Rationalize the denominator and simplify the expression.

$\displaystyle{\dfrac{7}{\sqrt{7}+4}=}$

35.

Rationalize the denominator and simplify the expression.

$\displaystyle{\dfrac{\sqrt{5}-13}{\sqrt{13}+3}=}$

36.

Rationalize the denominator and simplify the expression.

$\displaystyle{\dfrac{\sqrt{3}-14}{\sqrt{7}+10}=}$

Radical Expressions and Rational Exponents

37.

Without using a calculator, evaluate the expression.

$\displaystyle{ 125^{-\frac{2}{3}} = }$

38.

Without using a calculator, evaluate the expression.

$\displaystyle{ 8^{-\frac{5}{3}} = }$

39.

Without using a calculator, evaluate the expression.

$\displaystyle{ \left(\frac{1}{81}\right)^{-\frac{3}{4}} = }$

40.

Without using a calculator, evaluate the expression.

$\displaystyle{ \left(\frac{1}{9}\right)^{-\frac{3}{2}} = }$

41.

Without using a calculator, evaluate the expression.

$\displaystyle{ \sqrt[3]{125^{2}}= }$

42.

Without using a calculator, evaluate the expression.

$\displaystyle{ \sqrt[4]{81^{3}}= }$

43.

Without using a calculator, evaluate the expression.

$\displaystyle{ \sqrt[5]{1024}= }$

44.

Without using a calculator, evaluate the expression.

$\displaystyle{ \sqrt[3]{64}= }$

45.

Use rational exponents to write the expression.

$\displaystyle{\sqrt[5]{b}}$ =

46.

Use rational exponents to write the expression.

$\displaystyle{\sqrt{c}}$ =

47.

Use rational exponents to write the expression.

$\displaystyle{\sqrt[5]{8 x + 7}=}$

48.

Use rational exponents to write the expression.

$\displaystyle{\sqrt[4]{5 z + 1}=}$

49.

Convert the expression to radical notation.

$\displaystyle{{t^{\frac{2}{3}}}}$ =

50.

Convert the expression to radical notation.

$\displaystyle{{r^{\frac{4}{5}}}}$ =

51.

Convert the expression to radical notation.

$\displaystyle{{m^{\frac{5}{4}}}}$ =

52.

Convert the expression to radical notation.

$\displaystyle{{r^{\frac{2}{3}}}}$ =

53.

Convert the expression to radical notation.

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

54.

Convert the expression to radical notation.

$\displaystyle{{13^{\frac{1}{4}}b^{\frac{3}{4}}}}$ =

55.

Simplify the expression, answering with rational exponents and not radicals.

$\displaystyle{\sqrt[11]{c}\,\sqrt[11]{c}=}$

56.

Simplify the expression, answering with rational exponents and not radicals.

$\displaystyle{\sqrt[9]{x}\,\sqrt[9]{x}=}$

57.

Simplify the expression, answering with rational exponents and not radicals.

$\displaystyle{\sqrt[5]{32 z^{2}}=}$

58.

Simplify the expression, answering with rational exponents and not radicals.

$\displaystyle{\sqrt[3]{125 t^{5}}=}$

59.

Simplify the expression, answering with rational exponents and not radicals.

$\displaystyle{\frac{\sqrt{16 r}}{\sqrt[10]{r^{3}}}=}$

60.

Simplify the expression, answering with rational exponents and not radicals.

$\displaystyle{\frac{\sqrt{36 m}}{\sqrt[10]{m^{3}}}=}$

61.

Simplify the expression, answering with rational exponents and not radicals.

$\displaystyle{\sqrt{n} \cdot \sqrt[6]{n^{5}}=}$

62.

Simplify the expression, answering with rational exponents and not radicals.

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

Solving Radical Equations

63.

Solve the equation.

$\displaystyle{ {t} = {\sqrt{t-3}+5} }$

64.

Solve the equation.

$\displaystyle{ {t} = {\sqrt{t-1}+3} }$

65.

Solve the equation.

$\displaystyle{ {\sqrt{x+9}} = {\sqrt{x}+1} }$

66.

Solve the equation.

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

67.

Solve the equation.

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

68.

Solve the equation.

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

69.

Solve the equation.

$\displaystyle{ {r} = {\sqrt{r+4}+16} }$

70.

Solve the equation.

$\displaystyle{ {r} = {\sqrt{r+2}+88} }$

71.

Solve the equation.

$\displaystyle{ {\sqrt{52-t}} = {t+4} }$

72.

Solve the equation.

$\displaystyle{ {\sqrt{17-t}} = {t+3} }$

73.

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

c = \sqrt{a^{2} + b^{2}}

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

If a rectangular triangle has a hypothenuse of ${41\ {\rm ft}}$ and one leg is ${40\ {\rm ft}}$ long, how long is the third side of the triangle?

The third side of the triangle is long.

74.

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

c = \sqrt{a^{2} + b^{2}}

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

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.

75.

A pendulum has the length $L$ ft. The time period $T$ that it takes to once swing back and forth is $2$ s. Use the following formula to find its length.

T = 2 π \sqrt{\frac{L}{32}}

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

The pendulum is long.

76.

A pendulum has the length $L$ ft. The time period $T$ that it takes to once swing back and forth is $4$ s. Use the following formula to find its length.

T = 2 π \sqrt{\frac{L}{32}}

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

The pendulum is long.