Practice 2 Solutions

Solve. Graph the solution on a number line.

$\sqrt{x + 9} > \sqrt{3 x + 2}$

$\begin{array}{rcl} \begin{matrix} Res Restrictions & Solve \\ x + 9 \geq 0 & 3 x + 2 \geq 0 & {(\sqrt{x + 9})}^{2} > {(\sqrt{3 x + 2})}^{2} \\ x \geq - 9 & 3 x \geq - 2 & x + 9 > 3 x + 2 \\ x \geq - \frac{2}{3} & - 2 x > - 7 \\ x < \frac{7}{2} \\ - \frac{2}{3} \leq x < \frac{7}{2} \end{matrix} \end{array}$

$5 - 3 \sqrt{x} > - 4$

$\begin{array}{rcl} \begin{matrix} Res Restrictions & Solve \\ x \geq 0 & - 3 \sqrt{x} > - 9 \\ \sqrt{x} < 3 \\ {(\sqrt{x})}^{2} < 3^{2} \\ x < 9 \\ 0 \leq x < 9 \end{matrix} \end{array}$

$7 < \sqrt{2 x - 3}$

$\begin{array}{rcl} \begin{matrix} Restrictions & Solve \\ 2 x - 3 \geq 0 & {(7)}^{2} < {(\sqrt{2 x - 3})}^{2} \\ 2 x \geq 3 & 49 < 2 x - 3 \\ x \geq \frac{3}{2} & 52 < 2 x \\ 26 < x \\ x > 26 \end{matrix} \end{array}$

$\sqrt{2 x - 4} \geq 3 - \sqrt{2 x - 1}$

$\begin{matrix} Restrictions & Solve \\ 2 x - 4 \geq 0 & 2 x - 1 \geq 0 & {(\sqrt{2 x - 4})}^{2} \geq {(3 - \sqrt{2 x - 1})}^{2} \\ 2 x \geq 4 & 2 x \geq 1 & 2 x - 4 \geq 9 - 6 \sqrt{2 x - 1} + 2 x - 1 \\ x \geq 2 & x \geq \frac{1}{2} & - 12 \geq - 6 \sqrt{2 x - 1} \\ 2 \leq \sqrt{2 x - 1} \\ {(2)}^{2} \leq {(\sqrt{2 x - 1})}^{2} \\ 4 \leq 2 x - 1 \\ \frac{5}{2} \leq x \\ x \geq \frac{5}{2} \end{matrix}$

$\sqrt{6 x + 8} \leq \sqrt{9 x + 2}$

$\begin{matrix} Restrictions & Solve \\ 6 x + 8 \geq 0 & 9 x + 2 \geq 0 & {(\sqrt{6 x + 8})}^{2} \leq {(\sqrt{9 x + 2})}^{2} \\ 6 x \geq - 8 & 9 x \geq - 2 & 6 x + 8 \leq 9 x + 2 \\ x \geq - \frac{4}{3} & x \geq - \frac{2}{9} & 6 \leq 3 x \\ 2 \leq x \\ x \geq 2 \end{matrix}$

Note

You only need x ≥ 2 because it is also true for the restriction.

$\sqrt{x} + 8 \geq \sqrt{x - 6}$

$\begin{array}{rcl} \begin{matrix} Res Restrictions & Solve \\ x \geq 0 & x - 6 \geq 0 & {(\sqrt{x} + 8)}^{2} \geq {(\sqrt{x - 6})}^{2} \\ x \geq 6 & x + 16 \sqrt{x} + 64 \geq x - 6 \\ 16 \sqrt{x} \geq - 70 \end{matrix} \end{array}$

Note

The principal square root is positive. A positive number will always be greater than a negative number, so the only solution is the restriction.

$4 > \sqrt{6 - 2 x}$

$\begin{array}{rcl} \begin{matrix} Restrictions & Solve \\ 6 - 2 x \geq 0 & {(4)}^{2} > {(\sqrt{6 - 2 x})}^{2} \\ 6 \geq 2 x & 16 > 6 - 2 x \\ 3 \geq x & 10 > - 2 x \\ - 5 < x \\ - 5 < x \leq 3 \end{matrix} \end{array}$

$2 + \sqrt{\frac{1}{2} x + 2} < - 10$

$\begin{array}{rcl} \begin{matrix} Res Restrictions & Solve \\ \frac{1}{2} x + 2 \geq 0 & \begin{array}{rcl} \sqrt{\frac{1}{2} x + 2} \end{array} \begin{array}{rcl} ≮ \end{array} \begin{array}{rcl} - \end{array} \begin{array}{rcl} 12 \end{array} \\ \frac{1}{2} x \geq - 2 & No solution \\ x \geq - 4 \end{matrix} \end{array}$

No solution

Note

The principal square root is positive, and a positive number cannot be less than or equal to a negative number.

$3 + \sqrt{\frac{1}{2} x} < \sqrt{\frac{1}{2} x + 12}$

$\begin{array}{rcl} \begin{matrix} Restrictions & Solve \\ \frac{1}{2} x \geq 0 & \frac{1}{2} x + 12 \geq 0 & {(3 + \sqrt{\frac{1}{2} x})}^{2} < {(\sqrt{\frac{1}{2} x + 12})}^{2} \\ x \geq 0 & \frac{1}{2} x \geq - 12 & 9 + 6 \sqrt{\frac{1}{2} x} + \frac{1}{2} x < \frac{1}{2} x + 12 \\ x \geq - 24 & 6 \sqrt{\frac{1}{2} x} < 3 \\ \sqrt{\frac{1}{2} x} < \frac{1}{2} \\ {(\sqrt{\frac{1}{2} x})}^{2} < {(\frac{1}{2})}^{2} \\ (2) \frac{1}{2} x < \frac{1}{4} (2) \\ x < \frac{1}{2} \\ 0 \leq x < \frac{1}{2} \end{matrix} \end{array}$

$- 3 - \sqrt{\frac{1}{3} x + 9} \geq - 12$

$\begin{matrix} Res Restrictions & \begin{matrix} Solve \end{matrix} \\ \frac{1}{3} x + 9 \geq 0 & \begin{array}{rcl} - \sqrt{\frac{1}{3} x + 9} & \geq & - 9 \end{array} \\ \begin{array}{rcl} \frac{1}{3} x & \geq & - 9 \end{array} & \sqrt{\frac{1}{3} x + 9} \leq 9 \\ x \geq - 27 & {(\sqrt{\frac{1}{3} x + 9})}^{2} \leq {(9)}^{2} \\ \frac{1}{3} x + 9 \leq 81 \\ \frac{1}{3} x \leq 72 \\ x \leq 216 \\ - 27 \leq x \leq 216 \end{matrix}$