ORCCA Radical Functions and Equations Chapter Review

Section 13.6 Radical Functions and Equations Chapter Review

Subsection 13.6.1 Introduction to Radical Functions

In Section 13.1 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.

Example 13.6.1. The 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 13.1.27 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 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 \(\left[\frac{1}{3},\infty\right)\text{.}\)

Figure 13.6.2. Graph of \(f(x)=\sqrt{3x-1}+2\)

Example 13.6.3. The 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.

Example 13.6.4. The 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 13.1.27, the domain is \((-\infty,\infty)\text{.}\)

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

Figure 13.6.5. Graph of \(y=g(x)=-2\sqrt[3]{x+6}-1\)

Example 13.6.6. Other 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 13.1.27, 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 the graph the function. By the graph, the function's range is \((-\infty,8]\text{.}\)

Figure 13.6.7. Graph of \(y=h(x)=8-\frac{5}{2}\sqrt[4]{6-2x}\)

Subsection 13.6.2 Radical Expressions and Rational Exponents

In Section 13.2 we learned the rational exponent rule and added it to our list of exponent rules.

Example 13.6.8. Radical Expressions and Rational Exponents.

Simplify the expressions using Fact 13.2.1 or Fact 13.2.9.

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

Explanation

\begin{align*} 100^{\sfrac{1}{2}}\amp=\left(\sqrt{100}\right)\\ \amp=10 \end{align*}
\begin{align*} (-64)^{-\sfrac{1}{3}}\amp=\frac{1}{(-64)^{\sfrac{1}{3}}}\\ \amp=\frac{1}{\left(\sqrt[3]{(-64)}\right)}\\ \amp=\frac{1}{-4} \end{align*}
\begin{align*} -81^{\sfrac{3}{4}}\amp=-\left(\sqrt[4]{81}\right)^3\\ \amp=-3^3\\ \amp=-27 \end{align*}
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 13.6.9. More Expressions with Rational Exponents.

Use exponent properties in List 13.2.13 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

\begin{align*} 7z^{\sfrac{5}{9}}\amp=7\sqrt[9]{z^5} \end{align*}
\begin{align*} \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{align*}
\begin{align*} \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{align*}
\begin{align*} \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{align*}
\begin{align*} \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{align*}
\begin{align*} \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{align*}
\begin{alignat*}{1} 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{alignat*}
\begin{align*} -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{align*}

Subsection 13.6.3 Radical Expression Operations

In Section 13.3 we learned to apply properties previously learned in Section 9.1 to radicals with higher roots and to more complicated radical expressions. These properties include the following:

\begin{align*} \sqrt[m]{x^m}\amp=x\amp\text{where }m\text{ is odd}\\ \sqrt[n]{x^n}\amp=\abs{x}\amp\text{where }n\text{ is even}\\ \sqrt[n]{x\cdot y}\amp=\sqrt[n]{x}\cdot\sqrt[n]{y}\\ \sqrt[n]{\frac{x}{y}}\amp=\frac{\sqrt[n]{x}}{\sqrt[n]{y}}, \text{ where }y\ne0 \end{align*}

Example 13.6.10.

Simplify \(\sqrt{y^2-4y+4}\text{.}\)

Explanation

\begin{align*} \sqrt{y^2-4y+4}\amp=\sqrt{(y-2)^2}\\ \amp=\abs{y-2} \end{align*}

We used the property \(\sqrt[n]{x^n}=\abs{x}\text{ where }n\text{ is even}\text{.}\)

Example 13.6.11.

Simplify \(\sqrt[4]{48x^6y^9}\text{.}\) Assume all variables are positive.

Explanation

\begin{align*} \sqrt[4]{48x^6y^9}\amp=\sqrt[4]{16\cdot 3\cdot x^4\cdot x^2\cdot y^8\cdot y}\\ \amp=\sqrt[4]{2^4\cdot 3\cdot x^4\cdot x^2\cdot \left(y^2\right)^4\cdot y}\\ \amp=2xy^2\sqrt[4]{3x^2y} \end{align*}

We used the property \(\sqrt[n]{x^n}=\abs{x}\text{ where }m\text{ is even}\text{,}\) but absolute value signs were unnecessary since all variables are assumed to be positive.

Example 13.6.12.

Do multiplication: \(\sqrt[3]{4y}\cdot\sqrt[3]{10y^4}\)

Explanation

\begin{align*} \sqrt[3]{4y}\cdot\sqrt[3]{10y^4}\amp=\sqrt[3]{(4y)\cdot\left(10y^4\right)}\\ \amp=\sqrt[3]{40y^5}\\ \amp=\sqrt[3]{8\cdot 5\cdot y^3\cdot y^2}\\ \amp=\sqrt[3]{2^3\cdot 5\cdot y^3\cdot y^2}\\ \amp=2y\sqrt[3]{5y^2} \end{align*}

Example 13.6.13.

Do division: \(\frac{\sqrt[4]{5m^6n^{10}}}{\sqrt[4]{80m^{10}n}}\text{.}\) Assume all variables are positive.

Explanation

\begin{align*} \frac{\sqrt[4]{5a^5b^{12}}}{\sqrt[4]{80a^{17}b^3}}\amp=\sqrt[4]{\frac{5a^5b^{12}}{80a^{17}b^3}}\\ \amp=\sqrt[4]{\frac{1b^9}{16a^{12}}}\\ \amp=\sqrt[4]{\frac{b^8b}{2^4\left(a^3\right)^4}}\\ \amp=\frac{b^2\sqrt[4]{b}}{2a^3} \end{align*}

In addition, we reviewed how to add or subtract radical expressions from Section 9.1 and applied this concept to higher roots and more complicated radical expressions.

Example 13.6.14.

Simplify \(\sqrt[3]{128x^4}-3x\sqrt[3]{54x}\text{.}\)

Explanation

\begin{align*} \sqrt[3]{128x^4}-3x\sqrt[3]{54x}\amp=\sqrt[3]{64\cdot 2\cdot x^3\cdot x}-3x\sqrt[3]{27\cdot 2\cdot x}\\ \amp=\sqrt[3]{4^3\cdot 2\cdot x^3\cdot x}-3x\sqrt[3]{3^3\cdot 2\cdot x}\\ \amp=4x\sqrt[3]{2x}-9x\sqrt[3]{2x}\\ \amp=-5x\sqrt[3]{2x} \end{align*}

Subsection 13.6.4 More on Rationalizing the Denominator

In Section 13.4 we covered how to rationalize the denominator when it contains a single square root or a binomial.

Example 13.6.15. A Review of Rationalizing the Denominator.

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}}{\sqrt{9}}\\ \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\sqrt{9}}\\ \amp=\frac{\sqrt{15}}{5\cdot 3}\\ \amp=\frac{\sqrt{15}}{15} \end{align*}

Example 13.6.16. Rationalize 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*}

Subsection 13.6.5 Solving Radical Equations

In Section 13.5 we covered solving equations that contain a radical. We learned about extraneous solutions and the need to check our solutions.

Example 13.6.17. Solving Radical Equations.

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\\ r-6=0\amp\text{ or }r-13=0\\ r=6\amp\text{ or }r=13 \end{align*}

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

Substitute \(\substitute{6}\) into \(r=9+\sqrt{r+3}\text{,}\) and we have:

\begin{align*} r\amp=9+\sqrt{r+3}\\ \substitute{6}\amp\stackrel{?}{=}9+\sqrt{\substitute{6}+3}\\ 6\amp\stackrel{?}{=}9+\sqrt{9}\\ 6\amp\stackrel{?}{=}9+3\\ 6\amp\stackrel{\text{no}}{=}12 \end{align*}

It turns out \(6\) is an extraneous solution.

Next, we substitute \(\substitute{13}\) into \(r=9+\sqrt{r+3}\text{:}\)

\begin{align*} r\amp=9+\sqrt{r+3}\\ \substitute{13}\amp\stackrel{?}{=}9+\sqrt{\substitute{13}+3}\\ 13\amp\stackrel{?}{=}9+\sqrt{16}\\ 13\amp\stackrel{?}{=}9+4\\ 13\amp\stackrel{\checkmark}{=}13 \end{align*}

So, \(13\) is the solution.

The equation has one solution: \(13\text{.}\) The solution set is \(\{13\}\text{.}\)

Example 13.6.18. 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}\\ \frac{8}{2}\amp=\frac{2\sqrt{t}}{2}\\ 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 \(\{\text{ }\}\text{.}\)

Exercises 13.6.6 Exercises

Introduction to Radical Functions

1.

Find the domain of the function.

\(g(x)={\sqrt{5-x}}\)

2.

Find the domain of the function.

\(h(x)={\sqrt{2-x}}\)

3.

Find the domain of the function.

\(F(x)=\sqrt[3]{{5x+2}}\)

4.

Find the domain of the function.

\(F(x)=\sqrt[3]{{-3x-10}}\)

5.

Find the domain of the function.

\(G(x)=\sqrt[4]{{-2-2x}}\)

6.

Find the domain of the function.

\(H(x)=\sqrt[4]{{32-4x}}\)

7.

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.

After seconds since the release, the object would have traveled \(45\) feet.
After \(4.4\) seconds since the release, the object would have traveled feet.

8.

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.

After seconds since the release, the object would have traveled \(50\) feet.
After \(5\) seconds since the release, the object would have traveled feet.

9.

Find the distance between the points \((-16,14)\) and \((-61,42)\text{.}\)

10.

Find the distance between the points \((1,3)\) and \((-6,27)\text{.}\)

Radical Expressions and Rational Exponents

11.

Use a calculator to evaluate the expression as a decimal to four significant digits.

\(\displaystyle{\sqrt[7]{8^{4}}=}\)

12.

Use a calculator to evaluate the expression as a decimal to four significant digits.

\(\displaystyle{\sqrt[5]{10^{3}}=}\)

13.

Without using a calculator, evaluate the expression.

\(\displaystyle{ \sqrt[3]{{-{\frac{1}{125}}}}= }\) .

14.

Without using a calculator, evaluate the expression.

\(\displaystyle{ \sqrt[3]{{-{\frac{64}{125}}}}= }\) .

15.

Use rational exponents to write the expression.

\(\displaystyle{\sqrt{6 a + 7}=}\)

16.

Use rational exponents to write the expression.

\(\displaystyle{\sqrt[5]{3 b + 1}=}\)

17.

Convert the expression to radical notation.

\(\displaystyle{{10^{\frac{1}{6}}c^{\frac{5}{6}}}}\) =

18.

Convert the expression to radical notation.

\(\displaystyle{{18^{\frac{1}{4}}x^{\frac{3}{4}}}}\) =

19.

Convert \(z^{-\frac{2}{3}}\) to a radical expression.

20.

Convert \(t^{-\frac{5}{8}}\) to a radical expression.

21.

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

\(\displaystyle{\frac{\sqrt[3]{8 r}}{\sqrt[6]{r^{5}}}=}\)

22.

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

\(\displaystyle{\frac{\sqrt{36 m}}{\sqrt[6]{m^{5}}}=}\)

23.

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

\(\displaystyle{\sqrt{n} \cdot \sqrt[10]{n^{3}}=}\)

24.

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

\(\displaystyle{\sqrt[5]{a}\cdot\sqrt[10]{a^{3}}=}\)

25.

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

\(\displaystyle{\sqrt{x}\sqrt[9]{x}=}\)

26.

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

\(\displaystyle{\sqrt{a}\sqrt[10]{a}=}\)

Radical Expression Operations

27.

Simplify the following expression. Do not assume the variables take only positive values.

\(\sqrt{9a^2-12a+4}\)

28.

Simplify the following expression. Assume all variables are positive.

\(\sqrt{75a^{11}b^5}\)

29.

Do the multiplication and simplify the result. Assume all variables are positive.

\(\sqrt{24x^3y^2}\cdot\sqrt{2x^5y^3}\)

30.

Simplify.

\(\sqrt[3]{\frac{54}{8}}\)

31.

Simplify the expression.

\(\displaystyle{{10\sqrt{29}} - {14\sqrt{29}} + {16\sqrt{29}} =}\)

32.

Simplify the expression.

\(\displaystyle{{\sqrt{12}} + {\sqrt{48}} + {\sqrt{72}} + {\sqrt{18}} =}\)

33.

Simplify \(2\sqrt[4]{162}-\sqrt[4]{32}\text{.}\)

34.

Expand and simplify the expression.

\(\displaystyle{\left(8 - {\sqrt{5}}\right)\left(5 - 3 {\sqrt{5}}\right) =}\)

35.

Expand and simplify the expression.

\(\displaystyle{ \left(4+\sqrt{3}\right)^2 =}\)