Answer Key 10.2
[latexpage]
- \(\begin{array}{rrl}
\\ \\
\sqrt{x^2}&=&\sqrt{75} \\
x&=&\pm \sqrt{25\cdot 3} \\
x&=&\pm 5\sqrt{3}
\end{array}\) - \(\begin{array}{rrl}
\\
\sqrt[3]{x^3}&=&\sqrt[3]{-8} \\
x&=&-2
\end{array}\) - \(\begin{array}{rrrrl}
\\ \\ \\ \\
x^2&+&5&=&13 \\
&-&5&&-5 \\
\midrule
&&\sqrt{x^2}&=&\sqrt{8} \\
&&x&=&\pm \sqrt{4\cdot 2} \\
&&x&=&\pm 2\sqrt{2}
\end{array}\) - \(\begin{array}{rrrrl}
\\ \\ \\ \\ \\ \\ \\
4x^3&-&2&=&106 \\
&+&2&&+2 \\
\midrule
&&\dfrac{4x^3}{4}&=&\dfrac{108}{4} \\ \\
&&x^3&=&27 \\
&&\sqrt[3]{x^3}&=&\sqrt[3]{27} \\
&&x&=&3
\end{array}\) - \(\begin{array}{rrrrl}
\\ \\ \\ \\ \\ \\ \\ \\
3x^2&+&1&=&73 \\
&-&1&&-1 \\
\midrule
&&\dfrac{3x^2}{3}&=&\dfrac{72}{3} \\ \\
&&x^2&=&24 \\
&&\sqrt{x^2}&=&\pm \sqrt{24} \\
&&x&=&\pm \sqrt{4\cdot 6} \\
&&x&=&\pm 2\sqrt{6}
\end{array}\) - \(\sqrt{(x-4)^2}=\sqrt{49}\)
\(\begin{array}{rrrrrrr}
x&-&4&=&\pm 7 && \\
&&x&=&4 & \pm & 7 \\
&&x&=&11, & -3&
\end{array}\) - \(\sqrt[5]{(x+2)^5}=\sqrt[5]{-3^5}\)
\(\begin{array}{rrrrr}
x&+&2&=&-3 \\
&-&2&&-2 \\
\midrule
&&x&=&-5
\end{array}\) - \(\sqrt[4]{(5x+1)^4}=\pm \sqrt[4]{2^4}\)
\(\begin{array}{rrrrrrr}
5x&+&1&=&\pm &2& \\
&-&1&&-&1& \\
\midrule
&&5x&=&-1&\pm &2 \\ \\
&&x&=&-\dfrac{3}{5}&\text{or}&\dfrac{1}{5}
\end{array}\) - \(\begin{array}{rrrrrrr}
\\ \\ \\
(2x&+&5)^3&-&6&=&21 \\
&&&+&6&&+6 \\
\midrule
&&(2x&+&5)^3&=&27 \\
\end{array}\)\(\sqrt[3]{(2x+5)^3}=\sqrt[3]{27}\)
\(\begin{array}{rrrrr}
2x&+&5&=&3 \\
&-&5&&-5 \\
\midrule
&&2x&=&-2 \\
&&x&=&-1
\end{array}\) - \(\begin{array}{rrrrrrr}
\\ \\ \\
(2x&+&1)^2&+&3&=&21 \\
&&&-&3&&-3 \\
\midrule
&&(2x&+&1)^2&=&18
\end{array}\)\(\sqrt{(2x+1)^2}&=&\sqrt{18} \Rightarrow \sqrt{9\cdot 2}\Rightarrow \pm 3\sqrt{2}\)
\(\begin{array}{rrrrl}
2x&+&1&=&\pm 3\sqrt{2} \\
&-&1&&-1 \\
\midrule
&&\dfrac{2x}{2}&=&\dfrac{-1\pm 3\sqrt{2}}{2} \\ \\
&&x&=&\dfrac{-1\pm 3\sqrt{2}}{2}
\end{array}\) - \(\begin{array}{rrrrl}
\\ \\ \\ \\ \\ \\
(x&-&1)^{\frac{2}{3}}&=&2^4 \\
(x&-&1)^{\frac{2}{3}\cdot \frac{3}{2}}&=&2^{4\cdot \frac{3}{2}} \\
x&-&1&=&\pm 2^6 \\
&+&1&&+1 \\
\midrule
&&x&=&1 \pm 2^6 \\
&&x&=&65\text{ or }-63
\end{array}\) - \(\begin{array}{rrrrl}
\\ \\ \\ \\ \\
(x&-&1)^{\frac{3}{2}}&=&2^3 \\
(x&-&1)^{\frac{3}{2}\cdot \frac{2}{3}}&=&2^{3\cdot \frac{2}{3}} \\
x&-&1&=&2^2 \\
&+&1&=&+1 \\
\midrule
&&x&=&5
\end{array}\) - \(\begin{array}{rrlrl}
\\ \\ \\ \\ \\ \\
(2&-&\phantom{-}x)^{\frac{3}{2}}&=&\phantom{-}3^3 \\
(2&-&\phantom{-}x)^{\frac{3}{2}\cdot \frac{2}{3}}&=&\phantom{-}3^{3\cdot \frac{2}{3}} \\
2&-&\phantom{-}x&=&\phantom{-}3^2 \\
-2&&&&-2 \\
\midrule
&&-x&=&\phantom{-}7 \\
&&\phantom{-}x&=&-7
\end{array}\) - \(\begin{array}{rrlrl}
\\ \\ \\ \\ \\ \\ \\ \\
(2x&+&3)^{\frac{4}{3}}&=&2^4 \\
(2x&+&3)^{\frac{4}{3}\cdot \frac{3}{4}}&=&2^{4\cdot \frac{3}{4}} \\
2x&+&3&=&\pm 2^3 \\
&-&3&&-3 \\
\midrule
&&2x&=&5 \\
&&2x&=&-11 \\ \\
&&x&=&\dfrac{5}{2}, -\dfrac{11}{2}
\end{array}\) - \(\begin{array}{rrlrl}
\\ \\ \\ \\ \\ \\ \\ \\
(2x&-&3)^{\frac{2}{3}}&=&2^2 \\
(2x&-&3)^{\frac{2}{3}\cdot \frac{3}{2}}&=&2^{2\cdot \frac{3}{2}} \\
2x&-&3&=&\pm 2^3 \\
&+&3&&+3 \\
\midrule
&&2x&=&11 \\
&&2x&=&-5 \\ \\
&&x&=&\dfrac{11}{2}, -\dfrac{5}{2}
\end{array}\) - \(\begin{array}{rrlrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\
(3x&-&2)^{\frac{4}{5}}&=&2^4 \\
(3x&-&2)^{\frac{4}{5}\cdot \frac{5}{4}}&=&2^{4\cdot \frac{5}{4}} \\
3x&-&2&=&\pm 2^5 \\
&+&2&&+2 \\
\midrule
&&\dfrac{3x}{3}&=&\dfrac{34}{3} \\ \\
&&\dfrac{3x}{3}&=&\dfrac{-30}{3} \\ \\
&&x&=&\dfrac{34}{3}, -10
\end{array}\)
