ORCCA Radical Expressions and Equations Chapter Review

Section 6.5 Radical Expressions and Equations Chapter Review

Subsection 6.5.1 Square and \(n\)th Root Properties

In Section 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 \(\overbrace{r\cdot r\cdot\cdots\cdot r}^{n\text{ times}}=x\text{.}\)

There are two helpful properties for simplifying radicals.

List 6.5.1. Properties of Radicals for Multiplication and Division

If \(a\) and \(b\) are positive real numbers, and \(m\) is a positive [cross-reference to target(s) "item-integer-definition" missing or not unique], then we have the following properties:

Root of a Product Property: \(\displaystyle \sqrt[m]{a\cdot b} = \sqrt[m]{a} \cdot \sqrt[m]{b}\)
Root of a Quotient Property: \(\sqrt[m]{\frac{a}{b}} = \frac{\sqrt[m]{a}}{\sqrt[m]{b}}\) as long as \(b \neq 0\)

Checkpoint 6.5.2.

Simplify \(\sqrt{72}\text{.}\)
Simplify \(\sqrt[3]{72}\text{.}\)
Simplify \(\sqrt{\frac{72}{25}}\text{.}\)

Explanation.

\begin{equation*} \begin{aligned} \sqrt{72}\amp=\sqrt{4\cdot18}\\ \amp=\sqrt{4}\cdot\sqrt{18}\\ \amp=2\sqrt{18}\\ \amp=2\sqrt{9\cdot2}\\ \amp=2\sqrt{9}\cdot\sqrt{2}\\ \amp=2\cdot3\sqrt{2}\\ \amp=6\sqrt{2} \end{aligned} \end{equation*}
\begin{equation*} \begin{aligned} \sqrt[3]{72}\amp=\sqrt[3]{8\cdot9}\\ \amp=\sqrt[3]{8}\cdot\sqrt[3]{9}\\ \amp=2\sqrt[3]{9} \end{aligned} \end{equation*}
\begin{equation*} \begin{aligned} \sqrt{\frac{72}{25}}\amp=\frac{\sqrt{72}}{\sqrt{25}}\\ \amp=\frac{6\sqrt{2}}{5} \end{aligned} \end{equation*}

Subsection 6.5.2 Rationalizing the Denominator

In Section 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.

\(\displaystyle \frac{12}{\sqrt{3}}\)
\(\displaystyle \frac{\sqrt{5}}{\sqrt{75}}\)

Explanation.

\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 \(\frac{\sqrt{6}-\sqrt{5}}{\sqrt{3}+\sqrt{2}}\text{.}\)

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}}{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

In Section 3 we learned the rational exponent property and added it to our list of exponent properties.

Example 6.5.5. Radical Expressions and Rational Exponents.

Simplify the expressions using Fact 6.3.2 or Fact 6.3.9.

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

Explanation.

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

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

Explanation.

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

Subsection 6.5.4 Solving Radical Equations

In Section 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 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\wonder{=}-1-\sqrt{16}\\ \sqrt{25}\amp\wonder{=}-1-4\\ 5\amp\reject{=}-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.

\(\displaystyle{\sqrt{{{\frac{16}{49}}}}={}}\).

2.

Evaluate the following.

\(\displaystyle{\sqrt{{{\frac{25}{121}}}}={}}\).

3.

Evaluate the following.

\(-\sqrt{64}={}\).

4.

Evaluate the following.

\(-\sqrt{81}={}\).

Exercise Group.

5.

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

\(\displaystyle{ \frac{{\sqrt{125}}}{{\sqrt{5}}} =}\)

6.

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

\(\displaystyle{ \frac{{\sqrt{144}}}{{\sqrt{4}}} =}\)

7.

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

\(\displaystyle{ {\sqrt{8}} = }\)

8.

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

\(\displaystyle{ {\sqrt{98}} = }\)

Exercise Group.

9.

Simplify the expression.

\(3\sqrt{7} \cdot 5\sqrt{{25}} =\)

10.

Simplify the expression.

\(4\sqrt{13} \cdot 2\sqrt{{121}} =\)

11.

Simplify the expression.

\(\displaystyle{ \sqrt{\frac{5}{6}} \cdot \sqrt{\frac{1}{6}} =}\)

12.

Simplify the expression.

\(\displaystyle{ \sqrt{\frac{2}{7}} \cdot \sqrt{\frac{5}{7}} =}\)

Exercise Group.

13.

Simplify the expression.

\(\displaystyle{{17\sqrt{2}} - {18\sqrt{2}} =}\)

14.

Simplify the expression.

\(\displaystyle{{18\sqrt{11}} - {19\sqrt{11}} =}\)

15.

Simplify the expression.

\(\displaystyle{{\sqrt{63}} + {\sqrt{252}} =}\)

16.

Simplify the expression.

\(\displaystyle{{\sqrt{252}} + {\sqrt{63}} =}\)

Exercise Group.

17.

\({\sqrt[3]{64}}\text{.}\)

18.

\({\sqrt[3]{27}}\text{.}\)

19.

\({\sqrt[5]{-32}}\text{.}\)

20.

\({\sqrt[3]{-125}}\text{.}\)

21.

\({\sqrt[4]{-16}}\text{.}\)

22.

\({\sqrt[4]{-81}}\text{.}\)

23.

\({\sqrt[6]{128}}\text{.}\)

24.

\({\sqrt[6]{128}}\text{.}\)

25.

\({\sqrt[5]{\frac{7}{32}}}\text{.}\)

26.

\({\sqrt[3]{\frac{7}{27}}}\text{.}\)

27.

\({\sqrt[3]{\frac{135}{64}}}\text{.}\)