Practice 1 Solutions

The restrictions on the radicand sometimes create a lower or upper boundary when the solution is graphed on a number line.
The principal root can never be less than zero.

Solve. Graph the solution on a number line.

$\sqrt{2 x - 7} \leq \sqrt{x + 3}$

$\begin{array}{rcl} \begin{array}{cc} Res Restrictions & Solve \\ 2 x - 7 \geq 0 & x + 3 \geq 0 & {(\sqrt{2 x - 7})}^{2} \leq {(\sqrt{x + 3})}^{2} \\ 2 x \geq 7 & x \geq - 3 & 2 x - 7 \leq x + 3 \\ x \geq \frac{7}{2} & x \leq 10 \\ \frac{7}{2} \leq x \leq 10 \end{array} \end{array}$

$- \sqrt{5 x - 9} + 7 \leq - 4$

$\begin{array}{rcl} \begin{array}{cc} Res Restrictions & Solve \\ 5 x - 9 \geq 0 & - \sqrt{5 x - 9} \leq - 11 \\ 5 x \geq 9 & \sqrt{5 x - 9} \geq 11 \\ x \geq \frac{9}{5} & {(\sqrt{5 x - 9})}^{2} \geq 11^{2} \\ 5 x - 9 \geq 121 \\ 5 x \geq 130 \\ x \geq 26 \end{array} \end{array}$

Note

The restriction is not needed in the final answer because the inequality solution is more restrictive than the radicand inequality result.

$9 + 2 \sqrt{x - 1} \geq 1$

$\begin{array}{rcl} \begin{array}{ccc} Res Restrictions & Solve \\ x - 1 \geq 0 & 2 \sqrt{x - 1} \geq - 8 \\ x \geq 1 & \sqrt{x - 1} \geq - 4 \\ The principal root will always be \\ greater than a negative number . \end{array} \end{array}$

Note

Q: Why is a positive radical expression always greater than a negative number?

A: The principal square root is positive. A positive number will always be greater than a negative number.

Q: What does that tell you about the answer to this problem?

A: The only solution is the restriction.

To understand this solution, substitute the values {1, 5, 10}. You should notice that all values greater than or equal to 1 are true because the square root will always be greater than a negative number.

$\sqrt{x + 3} \leq 1 - \sqrt{x + 5}$

$\begin{array}{rcl} \begin{matrix} Res Restrictions & Solve \\ x + 3 \geq 0 & x + 5 \geq 0 & {(\sqrt{x + 3})}^{2} \leq {(1 - \sqrt{x + 5})}^{2} \\ x \geq - 3 & x \geq - 5 & x + 3 \leq 1 - 2 \sqrt{x + 5} + x + 5 \\ - 3 \leq - 2 \sqrt{x + 5} \\ \frac{3}{2} \geq \sqrt{x + 5} \\ {(\frac{3}{2})}^{2} \geq {(\sqrt{x + 5})}^{2} \\ \frac{9}{4} \geq x + 5 \\ - \frac{11}{4} \geq x \\ - 3 \leq x \leq - \frac{11}{4} \end{matrix} \end{array}$

$\sqrt{3 x + 9} < \sqrt{x + 23}$

$\begin{array}{rcl} \begin{matrix} Res Restrictions & Solve \\ 3 x + 9 \geq 0 & x + 23 \geq 0 & {(\sqrt{3 x + 9})}^{2} < {(\sqrt{x + 23})}^{2} \\ 3 x \geq - 9 & x \geq - 23 & 3 x + 9 < x + 23 \\ x \geq - 3 & 2 x < 14 \\ x < 7 \\ - 3 \leq x < 7 \end{matrix} \end{array}$

$- 10 - \sqrt{12 x + 7} \geq - 2$

$\begin{array}{rcl} \begin{array}{cc} R Restrictions & Solve \\ 12 x + 7 \geq 0 & - \sqrt{12 x + 7} \geq 8 \\ 12 x \geq - 7 & \sqrt{12 x + 7} ≰ - 8 \\ x \geq - \frac{7}{12} & No solution \end{array} \end{array}$

No solution

Note

The principal square root is positive, and a positive number cannot be less than or equal to a negative number. Compare this to problem 5. Notice that the direction of the inequality symbol is different.

$1 > \sqrt{3 - x}$

$\begin{array}{rcl} \begin{matrix} Res Restrictions & Solve \\ 3 - x \geq 0 & 1^{2} > {(\sqrt{3 - x})}^{2} \\ x \leq 3 & 1 > 3 - x \\ - 2 > - x \\ 2 < x \\ 2 < x \leq 3 \end{matrix} \end{array}$

$- 5 + \sqrt{\frac{3}{4} x - 2} \leq 2$

$\begin{array}{rcl} \begin{matrix} Restrictions & Solve \\ \frac{3}{4} x - 2 \geq 0 & \sqrt{\frac{3}{4} x - 2} \leq 7 \\ \frac{3}{4} x \geq 2 & {(\sqrt{\frac{3}{4} x - 2})}^{2} \leq {(7)}^{2} \\ x \geq \frac{8}{3} & \frac{3}{4} x - 2 \leq 49 \\ \frac{3}{4} x \leq 51 \\ (\frac{4}{3}) \frac{3}{4} x \leq 51 (\frac{4}{3}) \\ x \leq 68 \\ \frac{8}{3} \leq x \leq 68 \end{matrix} \end{array}$

$\sqrt{\frac{2}{5} x - 8} < \sqrt{\frac{1}{4} x - 1}$

$\begin{array}{rcl} \begin{matrix} Restrictions & Solve \\ \frac{2}{5} x - 8 \geq 0 & \frac{1}{4} x - 1 \geq 0 & {(\sqrt{\frac{2}{5} x - 8})}^{2} < {(\sqrt{\frac{1}{4} x - 1})}^{2} \\ \frac{2}{5} x \geq 8 & \frac{1}{4} x \geq 1 & \frac{2}{5} x - 8 < \frac{1}{4} x - 1 \\ (\frac{5}{2}) \frac{2}{5} x \geq 8 (\frac{5}{2}) & x \geq 4 & 20 (\frac{2}{5} x - 8 < \frac{1}{4} x - 1) \\ x \geq 20 & 8 x - 160 < 5 x - 20 \\ 3 x < 140 \\ x < \frac{140}{3} \\ 20 \leq x < \frac{140}{3} \end{matrix} \end{array}$

$2 + \sqrt{3 x - 6} < \sqrt{3 x + 1}$

$\begin{array}{rcl} \begin{matrix} Res Restrictions & Solve \\ 3 x - 6 \geq 0 & 3 x + 1 \geq 0 & {(2 + \sqrt{3 x - 6})}^{2} < {(\sqrt{3 x + 1})}^{2} \\ 3 x \geq 6 & 3 x \geq - 1 & 4 + 4 \sqrt{3 x - 6} + 3 x - 6 < 3 x + 1 \\ x \geq 2 & x \geq - \frac{1}{3} & 4 \sqrt{3 x - 6} < 3 \\ \sqrt{3 x - 6} < \frac{3}{4} \\ {(\sqrt{3 x - 6})}^{2} < {(\frac{3}{4})}^{2} \\ 3 x - 6 < \frac{9}{16} \\ 3 x < 6 \frac{9}{16} \\ 3 x < \frac{105}{16} \\ (\frac{1}{3}) 3 x < \frac{105}{16} (\frac{1}{3}) \\ x < \frac{35}{16} \\ 2 \leq x < \frac{35}{16} \end{matrix} \end{array}$