Answer Key 10.2

Terrance Berg

Answer Key 10.2

$\begin{array}{rrl} \sqrt{x^{2}} & = & \sqrt{75} \\ x & = & \pm \sqrt{25 \cdot 3} \\ x & = & \pm 5 \sqrt{3} \end{array}$
$\begin{array}{rrl} \sqrt[3]{x^{3}} & = & \sqrt[3]{- 8} \\ x & = & - 2 \end{array}$
$\begin{array}{rrrrl} x^{2} & + & 5 & = & 13 \\ - & 5 & - 5 \\ \sqrt{x^{2}} & = & \sqrt{8} \\ x & = & \pm \sqrt{4 \cdot 2} \\ x & = & \pm 2 \sqrt{2} \end{array}$
$\begin{array}{rrrrl} 4 x^{3} & - & 2 & = & 106 \\ + & 2 & + 2 \\ \frac{4 x^{3}}{4} & = & \frac{108}{4} \\ x^{3} & = & 27 \\ \sqrt[3]{x^{3}} & = & \sqrt[3]{27} \\ x & = & 3 \end{array}$
$\begin{array}{rrrrl} 3 x^{2} & + & 1 & = & 73 \\ - & 1 & - 1 \\ \frac{3 x^{2}}{3} & = & \frac{72}{3} \\ x^{2} & = & 24 \\ \sqrt{x^{2}} & = & \pm \sqrt{24} \\ x & = & \pm \sqrt{4 \cdot 6} \\ x & = & \pm 2 \sqrt{6} \end{array}$
$\sqrt{(x - 4)^{2}} = \sqrt{49}$
$\begin{array}{rrrrrrr} x & - & 4 & = & \pm 7 \\ x & = & 4 & \pm & 7 \\ x & = & 11, & - 3 \end{array}$
$\sqrt[5]{(x + 2)^{5}} = \sqrt[5]{- 3^{5}}$
$\begin{array}{rrrrr} x & + & 2 & = & - 3 \\ - & 2 & - 2 \\ x & = & - 5 \end{array}$
$\sqrt[4]{(5 x + 1)^{4}} = \pm \sqrt[4]{2^{4}}$
$\begin{array}{rrrrrrr} 5 x & + & 1 & = & \pm & 2 \\ - & 1 & - & 1 \\ 5 x & = & - 1 & \pm & 2 \\ x & = & - \frac{3}{5} & or & \frac{1}{5} \end{array}$
$\begin{array}{rrrrrrr} (2 x & + & 5)^{3} & - & 6 & = & 21 \\ + & 6 & + 6 \\ (2 x & + & 5)^{3} & = & 27 \end{array}$
$\sqrt[3]{(2 x + 5)^{3}} = \sqrt[3]{27}$

$\begin{array}{rrrrr} 2 x & + & 5 & = & 3 \\ - & 5 & - 5 \\ 2 x & = & - 2 \\ x & = & - 1 \end{array}$
$\begin{array}{rrrrrrr} (2 x & + & 1)^{2} & + & 3 & = & 21 \\ - & 3 & - 3 \\ (2 x & + & 1)^{2} & = & 18 \end{array}$
$Math input error$

$\begin{array}{rrrrl} 2 x & + & 1 & = & \pm 3 \sqrt{2} \\ - & 1 & - 1 \\ \frac{2 x}{2} & = & \frac{- 1 \pm 3 \sqrt{2}}{2} \\ x & = & \frac{- 1 \pm 3 \sqrt{2}}{2} \end{array}$
$\begin{array}{rrrrl} (x & - & 1)^{\frac{2}{3}} & = & 2^{4} \\ (x & - & 1)^{\frac{2}{3} \cdot \frac{3}{2}} & = & 2^{4 \cdot \frac{3}{2}} \\ x & - & 1 & = & \pm 2^{6} \\ + & 1 & + 1 \\ x & = & 1 \pm 2^{6} \\ x & = & 65 or - 63 \end{array}$
$\begin{array}{rrrrl} (x & - & 1)^{\frac{3}{2}} & = & 2^{3} \\ (x & - & 1)^{\frac{3}{2} \cdot \frac{2}{3}} & = & 2^{3 \cdot \frac{2}{3}} \\ x & - & 1 & = & 2^{2} \\ + & 1 & = & + 1 \\ x & = & 5 \end{array}$
$\begin{array}{rrlrl} (2 & - & x)^{\frac{3}{2}} & = & 3^{3} \\ (2 & - & x)^{\frac{3}{2} \cdot \frac{2}{3}} & = & 3^{3 \cdot \frac{2}{3}} \\ 2 & - & x & = & 3^{2} \\ - 2 & - 2 \\ - x & = & 7 \\ x & = & - 7 \end{array}$
$\begin{array}{rrlrl} (2 x & + & 3)^{\frac{4}{3}} & = & 2^{4} \\ (2 x & + & 3)^{\frac{4}{3} \cdot \frac{3}{4}} & = & 2^{4 \cdot \frac{3}{4}} \\ 2 x & + & 3 & = & \pm 2^{3} \\ - & 3 & - 3 \\ 2 x & = & 5 \\ 2 x & = & - 11 \\ x & = & \frac{5}{2}, - \frac{11}{2} \end{array}$
$\begin{array}{rrlrl} (2 x & - & 3)^{\frac{2}{3}} & = & 2^{2} \\ (2 x & - & 3)^{\frac{2}{3} \cdot \frac{3}{2}} & = & 2^{2 \cdot \frac{3}{2}} \\ 2 x & - & 3 & = & \pm 2^{3} \\ + & 3 & + 3 \\ 2 x & = & 11 \\ 2 x & = & - 5 \\ x & = & \frac{11}{2}, - \frac{5}{2} \end{array}$
$\begin{array}{rrlrl} (3 x & - & 2)^{\frac{4}{5}} & = & 2^{4} \\ (3 x & - & 2)^{\frac{4}{5} \cdot \frac{5}{4}} & = & 2^{4 \cdot \frac{5}{4}} \\ 3 x & - & 2 & = & \pm 2^{5} \\ + & 2 & + 2 \\ \frac{3 x}{3} & = & \frac{34}{3} \\ \frac{3 x}{3} & = & \frac{- 30}{3} \\ x & = & \frac{34}{3}, - 10 \end{array}$

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Intermediate Algebra (Convert to MathJax) Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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