# Answer Key 10.1

1. $\phantom{a}$
$\begin{array}[t]{rrrrr} \sqrt{2x+3}&-&3&=&0 \\ &+&3&&+3 \\ \hline (\sqrt{2x+3})^2&&&=&\phantom{+}(3)^2 \\ \\ 2x&+&3&=&9 \\ &-&3&&-3 \\ \hline &&\dfrac{2x}{2}&=&\dfrac{6}{2} \\ \\ &&x&=&3 \end{array}$
2. $\phantom{a}$
$\begin{array}[t]{rrrrr} \sqrt{5x+1}&-&4&=&0 \\ &+&4&=&+4 \\ \hline (\sqrt{5x+1})^2&&&=&(4)^2 \\ \\ 5x&+&1&=&16 \\ &-&1&&-1 \\ \hline &&\dfrac{5x}{5}&=&\dfrac{15}{5} \\ \\ &&x&=&3 \end{array}$
3. $\phantom{a}$
$\begin{array}[t]{rrrrrrrrr} \sqrt{6x-5}&-&x&=&0&&&& \\ &+&x&&+x&&&& \\ \hline (\sqrt{6x-5})^2&&&=&(x)^2&&&& \\ \\ 6x&-&5&=&x^2&&&& \\ -6x&+&5&&&-&6x&+&5 \\ \hline &&0&=&x^2&-&6x&+&5 \\ &&0&=&(x&-&5)(x&-&1) \\ \\ &&x&=&5,&1&&& \end{array}$
4. $(\sqrt{7x+8})^2=(x)^2$
$\begin{array}[t]{rrrrrrrrr}7x&+&8&=&x^2&&&& \\ -7x&-&8&&&-&7x&-&8 \\ \hline &&0&=&x^2&-&7x&-&8 \\ &&0&=&(x&-&8)(x&+&1) \\ \\ &&x&=&-1,&8&&& \end{array}$
5. $(\sqrt{3+x})^2=(\sqrt{6x+13})^2$
$\begin{array}[t]{rrrrrrr} 3&+&x&=&6x&+&13 \\ -3&-&6x&&-6x&-&3 \\ \hline &&\dfrac{-5x}{-5}&=&\dfrac{10}{-5}&& \\ \\ &&x&=&-2&& \end{array}$
6. $(\sqrt{x-1})^2=(\sqrt{7-x})^2$
$\begin{array}{rrrrrrr} x&-&1&=&7&-&x \\ +x&+&1&&+1&+&x \\ \hline &&\dfrac{2x}{2}&=&\dfrac{8}{2}&& \\ \\ &&x&=&4&& \end{array}$
7. $(\sqrt[3]{3-3x})^3=(\sqrt[3]{2x-5})^3$
$\begin{array}{rrrrrrr} 3&-&3x&=&2x&-&5 \\ -3&-&2x&&-2x&-&3 \\ \hline &&\dfrac{-5x}{-5}&=&\dfrac{-8}{-5}&& \\ \\ &&x&=&\dfrac{8}{5}&& \end{array}$
8. $(\sqrt[4]{3x-2})^4=(\sqrt[4]{x+4})^4$
$\begin{array}{rrrrrrr} 3x&-&2&=&x&+&4 \\ -x&+&2&&-x&+&2 \\ \hline &&\dfrac{2x}{2}&=&\dfrac{6}{2}&& \\ \\ &&x&=&3&& \end{array}$
9. $(\sqrt{x+7})^2\ge (2)^2$
$\begin{array}{rrrrr} x&+&7&\ge &4 \\ &-&7&&-7 \\ \hline &&x&\ge &-3 \end{array}$
10. $(\sqrt{x-2})^2\le (4)^2$
$\begin{array}{rrrrr} x&-&2&\le &16 \\ &+&2&&+2 \\ \hline &&x&\le &18 \end{array}$
11. $(3)^2 < (\sqrt{3x+6})^2 \le (6)^2$ $\phantom{a}$ $\begin{array}[t]{rrrcrrr} 9&<&3x&+&6&\le &36 \\ -6&&&-&6&&-6 \\ \hline \dfrac{3}{3}&<&&\dfrac{3x}{3}&&\le &\dfrac{30}{3} \\ 1&<&&x&&\le &10 \end{array}$
12. $(0)^2 < (\sqrt{x+5})^2 < (5)^2$ $\phantom{a}$ $\begin{array}[t]{rrrcrrr} 0&<&x&+&5&<&25 \\ -5&&&-&5&&-5 \\ \hline -5&<&&x&&<&20 \end{array}$

## License

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