Answer Key 10.1
[latexpage]
- \(\begin{array}{rrrrr}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\
\sqrt{2x+3}&-&3&=&0 \\
&+&3&&+3 \\
\midrule
(\sqrt{2x+3})^2&&&=&\phantom{+}(3)^2 \\ \\
2x&+&3&=&9 \\
&-&3&&-3 \\
\midrule
&&\dfrac{2x}{2}&=&\dfrac{6}{2} \\ \\
&&x&=&3
\end{array}\) - \(\begin{array}{rrrrr}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\
\sqrt{5x+1}&-&4&=&0 \\
&+&4&=&+4 \\
\midrule
(\sqrt{5x+1})^2&&&=&(4)^2 \\ \\
5x&+&1&=&16 \\
&-&1&&-1 \\
\midrule
&&\dfrac{5x}{5}&=&\dfrac{15}{5} \\ \\
&&x&=&3
\end{array}\) - \(\begin{array}{rrrrrrrrr}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\
\sqrt{6x-5}&-&x&=&0&&&& \\
&+&x&&+x&&&& \\
\midrule
(\sqrt{6x-5})^2&&&=&(x)^2&&&& \\ \\
6x&-&5&=&x^2&&&& \\
-6x&+&5&&&-&6x&+&5 \\
\midrule
&&0&=&x^2&-&6x&+&5 \\
&&0&=&(x&-&5)(x&-&1) \\ \\
&&x&=&5,&1&&&
\end{array}\) - \((\sqrt{7x+8})^2=(x)^2\)
\(\begin{array}{rrrrrrrrr}
7x&+&8&=&x^2&&&& \\
-7x&-&8&&&-&7x&-&8 \\
\midrule
&&0&=&x^2&-&7x&-&8 \\
&&0&=&(x&-&8)(x&+&1) \\ \\
&&x&=&-1,&8&&&
\end{array}\) - \((\sqrt{3+x})^2=(\sqrt{6x+13})^2\)
\(\begin{array}{rrrrrrr}
3&+&x&=&6x&+&13 \\
-3&-&6x&&-6x&-&3 \\
\midrule
&&\dfrac{-5x}{-5}&=&\dfrac{10}{-5}&& \\ \\
&&x&=&-2&&
\end{array}\) - \((\sqrt{x-1})^2=(\sqrt{7-x})^2\)
\(\begin{array}{rrrrrrr}
x&-&1&=&7&-&x \\
+x&+&1&&+1&+&x \\
\midrule
&&\dfrac{2x}{2}&=&\dfrac{8}{2}&& \\ \\
&&x&=&4&&
\end{array}\) - \((\sqrt[3]{3-3x})^3=(\sqrt[3]{2x-5})^3\)
\(\begin{array}{rrrrrrr}
3&-&3x&=&2x&-&5 \\
-3&-&2x&&-2x&-&3 \\
\midrule
&&\dfrac{-5x}{-5}&=&\dfrac{-8}{-5}&& \\ \\
&&x&=&\dfrac{8}{5}&&
\end{array}\) - \((\sqrt[4]{3x-2})^4=(\sqrt[4]{x+4})^4\)
\(\begin{array}{rrrrrrr}
3x&-&2&=&x&+&4 \\
-x&+&2&&-x&+&2 \\
\midrule
&&\dfrac{2x}{2}&=&\dfrac{6}{2}&& \\ \\
&&x&=&3&&
\end{array}\) - \((\sqrt{x+7})^2\ge (2)^2\)
\(\begin{array}{rrrrr}
x&+&7&\ge &4 \\
&-&7&&-7 \\
\midrule
&&x&\ge &-3
\end{array}\) - \((\sqrt{x-2})^2\le (4)^2\)
\(\begin{array}{rrrrr}
x&-&2&\le &16 \\
&+&2&&+2 \\
\midrule
&&x&\le &18
\end{array}\) - \((3)^2 < (\sqrt{3x+6})^2 \le (6)^2\)
\(\begin{array}{rrrcrrr}
9&<&3x&+&6&\le &36 \\
-6&&&-&6&&-6 \\
\midrule
\dfrac{3}{3}&<&&\dfrac{3x}{3}&&\le &\dfrac{30}{3} \\ \\
1&<&&x&&\le &10
\end{array}\) - \((0)^2 < (\sqrt{x+5})^2 < (5)^2\)
\(\begin{array}{rrrcrrr}
0&<&x&+&5&<&25 \\
-5&&&-&5&&-5 \\
\midrule
-5&<&&x&&<&20
\end{array}\)
