Answer Key 10.2

  1. \begin{array}{rrl} \\ \\ \sqrt{x^2}&=&\sqrt{75} \\ x&=&\pm \sqrt{25\cdot 3} \\ x&=&\pm 5\sqrt{3} \end{array}
  2. \begin{array}{rrl} \\ \sqrt[3]{x^3}&=&\sqrt[3]{-8} \\ x&=&-2 \end{array}
  3. \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}
  4. \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}
  5. \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}
  6. \sqrt{(x-4)^2}=\sqrt{49}

    \begin{array}{rrrrrrr} x&-&4&=&\pm 7 && \\ &&x&=&4 & \pm & 7  \\ &&x&=&11, & -3& \end{array}

  7. \sqrt[5]{(x+2)^5}=\sqrt[5]{-3^5}

    \begin{array}{rrrrr} x&+&2&=&-3 \\ &-&2&&-2 \\ \midrule &&x&=&-5 \end{array}

  8. \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}

  9. \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}

  10. \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}

  11. \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}
  12. \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}
  13. \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}
  14. \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}
  15. \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}
  16. \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}

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Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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