Test 6 (Lessons 11–12): Radical Expressions (using real numbers) Solutions

1. $\sqrt{3}\left(2\sqrt{3}–\sqrt{5}\right)$

$2\sqrt{9}–\sqrt{15}$

$6–\sqrt{15}$

2. $\sqrt[5]{15a^{11}{b}^4{c}^{20}}$

$$\begin{gathered} {15}^{\frac{1}{5}}{a}^{\frac{11}{5}}{b}^{\frac{4}{5}}{c}^{\frac{20}{5}} \\ {15}^{\frac{1}{5}}{a}^{2\frac{1}{5}}{b}^{\frac{4}{5}}{c}^4 \end{gathered}$$

$a^2{c}^4 \sqrt[5]{15a{b}^4}$

3. ${\left(3x^7{y}^{10}{z}^9\right)}^{\frac{1}{2}}$

$$\begin{gathered} 3^{\frac{1}{2}}{x}^{\frac{7}{2}}{y}^{\frac{10}{2}}{z}^{\frac{9}{2}} \\ \sqrt{3}{x}^{3\frac{1}{2} }{y}^5 z^{4\frac{1}{2}} \end{gathered}$$

$\left|x^3{y}^5\right|z^4\sqrt{3xz}$

4. $\left(\sqrt[3]{9x^{2}y{z}^4}\right)\left(\sqrt[3]{12xy{z}^2}\right)$

$$\begin{gathered} \sqrt[3]{3^2·3·2^2{x}^3{y}^2{z}^6} \\ {2}^{\frac{2}{3}}·3^{\frac{3}{3}}{x}^{\frac{3}{3}}{y}^{\frac{2}{3}}{z}^{\frac{6}{3}} \end{gathered}$$

$3x{z}^2 \sqrt[3]{4y^2}$

5. The area of a right triangle is 36 square inches. The base of the triangle is $4\sqrt{6}$ inches long. What is the height of the triangle? Show your work.

$$\begin{gathered} A=36, b=4\sqrt{6} \\ A=\frac{1}{2}bh \\ 36=\frac{1}{2}\left(4\sqrt{6}\right)h \\ 36=\left(2\sqrt{6}\right)h \\ h=\frac{36}{2\sqrt{6}}=\frac{18}{\sqrt{6}} \\ h=\frac{18}{\sqrt{6}}\left(\frac{\sqrt{6}}{\sqrt{6}}\right)=\frac{18\sqrt{6}}{6} \\ h=3\sqrt{6} \end{gathered}$$

The height of the triangle is $3\sqrt{6}$inches.

6. $\frac{5\sqrt{2}}{\sqrt{12}}$

$$\begin{gathered} \frac{5\sqrt{2}}{\sqrt{2·6}}=\frac{5}{\sqrt{6}} \\ \frac{5}{\sqrt{6}}\left(\frac{\sqrt{6}}{\sqrt{6}}\right) \end{gathered}$$

$\frac{5\sqrt{6}}{6}$

7. $\frac{4}{2+\sqrt{11}}$

$$\begin{gathered} \frac{4}{2+\sqrt{11}}\left(\frac{2–\sqrt{11}}{2–\sqrt{11}}\right) \\ \frac{8–4\sqrt{11}}{4–11} \end{gathered}$$

$\frac{8–4\sqrt{11}}{–7} \mathrm{or} \frac{–8+4\sqrt{11}}{7}$

8. $7\sqrt{8}\left(3\sqrt{2}+1\right)–2\sqrt{2}$

$$\begin{gathered} 21\sqrt{16}+7\sqrt{8}–2\sqrt{2} \\ 21\sqrt{4^2}+7\sqrt{2·2^2}–2\sqrt{2} \\ 84+14\sqrt{2}–2\sqrt{2} \end{gathered}$$

$84+12\sqrt{2}$

9. $–\sqrt{15}\left(4\sqrt{5}–8\sqrt{3}\right)$

$–4\sqrt{3·5^2}+8\sqrt{3^2·5}$

$–20\sqrt{3}+24\sqrt{5}$

10. The area of a rectangle is $29–7\sqrt{17}$ square yards. The length of the rectangle is $4–\sqrt{17}$ yards. Find the width of the rectangle. Show your work.

$$\begin{gathered} A=lw; w=\frac{A}{l} \\ A=29–7\sqrt{17}, l=4–\sqrt{17} \\ w=\frac{29–7\sqrt{17}}{4–\sqrt{17}} \\ \frac{29–7\sqrt{17}}{4–\sqrt{17}}\left(\frac{4+\sqrt{17}}{4+\sqrt{17}}\right) \\ \frac{116+29\sqrt{17}–28\sqrt{17}–7\left(17\right)}{16–17} \\ \frac{116–\sqrt{17}–119}{–1} \end{gathered}$$

$w=3+\sqrt{17} \mathrm{yards}$