Answer Key 10.1

  1. \begin{array}{rrrrr} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \sqrt{2x+3}&-&3&=&0 \\ &+&3&&+3 \\ \midrule (\sqrt{2x+3})^2&&&=&\phantom{+}(3)^2 \\ \\ 2x&+&3&=&9 \\ &-&3&&-3 \\ \midrule &&\dfrac{2x}{2}&=&\dfrac{6}{2} \\ \\ &&x&=&3 \end{array}
  2. \begin{array}{rrrrr} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \sqrt{5x+1}&-&4&=&0 \\ &+&4&=&+4 \\ \midrule (\sqrt{5x+1})^2&&&=&(4)^2 \\ \\ 5x&+&1&=&16 \\ &-&1&&-1 \\ \midrule &&\dfrac{5x}{5}&=&\dfrac{15}{5} \\ \\ &&x&=&3 \end{array}
  3. \begin{array}{rrrrrrrrr} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \sqrt{6x-5}&-&x&=&0&&&& \\ &+&x&&+x&&&& \\ \midrule (\sqrt{6x-5})^2&&&=&(x)^2&&&& \\ \\ 6x&-&5&=&x^2&&&& \\ -6x&+&5&&&-&6x&+&5 \\ \midrule &&0&=&x^2&-&6x&+&5 \\ &&0&=&(x&-&5)(x&-&1) \\ \\ &&x&=&5,&1&&& \end{array}
  4. (\sqrt{7x+8})^2=(x)^2

    \begin{array}{rrrrrrrrr} 7x&+&8&=&x^2&&&& \\ -7x&-&8&&&-&7x&-&8 \\ \midrule &&0&=&x^2&-&7x&-&8 \\ &&0&=&(x&-&8)(x&+&1) \\ \\ &&x&=&-1,&8&&& \end{array}

  5. (\sqrt{3+x})^2=(\sqrt{6x+13})^2

    \begin{array}{rrrrrrr} 3&+&x&=&6x&+&13 \\ -3&-&6x&&-6x&-&3 \\ \midrule &&\dfrac{-5x}{-5}&=&\dfrac{10}{-5}&& \\ \\ &&x&=&-2&& \end{array}

  6. (\sqrt{x-1})^2=(\sqrt{7-x})^2

    \begin{array}{rrrrrrr} x&-&1&=&7&-&x \\ +x&+&1&&+1&+&x \\ \midrule &&\dfrac{2x}{2}&=&\dfrac{8}{2}&& \\ \\ &&x&=&4&& \end{array}

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

    \begin{array}{rrrrrrr} 3&-&3x&=&2x&-&5 \\ -3&-&2x&&-2x&-&3 \\ \midrule &&\dfrac{-5x}{-5}&=&\dfrac{-8}{-5}&& \\ \\ &&x&=&\dfrac{8}{5}&& \end{array}

  8. (\sqrt[4]{3x-2})^4=(\sqrt[4]{x+4})^4

    \begin{array}{rrrrrrr} 3x&-&2&=&x&+&4 \\ -x&+&2&&-x&+&2 \\ \midrule &&\dfrac{2x}{2}&=&\dfrac{6}{2}&& \\ \\ &&x&=&3&& \end{array}

  9. (\sqrt{x+7})^2\ge (2)^2

    \begin{array}{rrrrr} x&+&7&\ge &4 \\ &-&7&&-7 \\ \midrule &&x&\ge &-3 \end{array}

  10. (\sqrt{x-2})^2\le (4)^2

    \begin{array}{rrrrr} x&-&2&\le &16 \\ &+&2&&+2 \\ \midrule &&x&\le &18 \end{array}

  11. (3)^2 < (\sqrt{3x+6})^2 \le (6)^2

    \begin{array}{rrrcrrr} 9&<&3x&+&6&\le &36 \\ -6&&&-&6&&-6 \\ \midrule \dfrac{3}{3}&<&&\dfrac{3x}{3}&&\le &\dfrac{30}{3} \\ \\ 1&<&&x&&\le &10 \end{array}

  12. (0)^2 < (\sqrt{x+5})^2 < (5)^2

    \begin{array}{rrrcrrr} 0&<&x&+&5&<&25 \\ -5&&&-&5&&-5 \\ \midrule -5&<&&x&&<&20 \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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