Test 7 (Lessons 13-14): Radical Equations and Inequalities Solutions

Solve.

$\sqrt{x - 3} + 6 = 10$

$\begin{array}{rcl} \sqrt{x - 3} & = & 4 \\ {(\sqrt{x - 3})}^{2} & = & {(4)}^{2} \\ x - 3 & = & 16 \end{array}$

x = 19

${(x + 6)}^{\frac{1}{4}} = {(4 x + 7)}^{\frac{1}{4}}$

$\begin{array}{rcl} {({(x + 6)}^{\frac{1}{4}})}^{4} & = & {({(4 x + 7)}^{\frac{1}{4}})}^{4} \\ x + 6 & = & 4 x + 7 \\ - 1 & = & 3 x \end{array}$

$x = - \frac{1}{3}$

$\sqrt{x + 4} = 3 - \sqrt{x + 1}$

$\begin{array}{rcl} {(\sqrt{x + 4})}^{2} & = & {(3 - \sqrt{x + 1})}^{2} \\ x + 4 & = & 9 - 6 \sqrt{x + 1} + x + 1 \\ x + 4 & = & x + 10 - 6 \sqrt{x + 1} \\ - 6 & = & - 6 \sqrt{x + 1} \\ 1 & = & \sqrt{x + 1} \\ {(1)}^{2} & = & {(\sqrt{x + 1})}^{2} \\ 1 & = & x + 1 \end{array}$

x = 0

$\sqrt{x + 12} - \sqrt{x} = 5$

$\begin{array}{rcl} \sqrt{x + 12} & = & 5 + \sqrt{x} \\ {(\sqrt{x + 12})}^{2} & = & {(5 + \sqrt{x})}^{2} \\ x + 12 & = & 25 + 10 \sqrt{x} + x \\ - 13 & = & 10 \sqrt{x} \\ - \frac{13}{10} & = & \sqrt{x} \end{array}$

no real solutions

Note

$\sqrt{x}$ has no real solutions because taking the square root does not result in a negative number.

${(x + 5)}^{\frac{2}{3}} - 7 = - 3$

$\begin{array}{rcl} {(x + 5)}^{\frac{2}{3}} = 4 {({(x + 5)}^{\frac{2}{3}})}^{\frac{3}{2}} = {(4)}^{\frac{3}{2}} {(4)}^{\frac{3}{2}} = {(4^{\frac{1}{2}})}^{3} = 2^{3} | x + 5 | = 8 \\ \begin{matrix} case 1 & case 2 \\ x + 5 = 8 & - (x + 5) = 8 \\ x = 3 & x + 5 = - 8 \\ x = - 13 \end{matrix} \end{array}$

x = –13, 3

Note

When the numerator of a rational exponent is even, remember to use absolute value bars.

$\sqrt{x - 9} = 9 - \sqrt{x}$

$\begin{array}{rcl} {(\sqrt{x - 9})}^{2} & = & {(9 - \sqrt{x})}^{2} \\ x - 9 & = & 81 - 18 \sqrt{x} + x \\ - 90 & = & - 18 \sqrt{x} \\ 5 & = & \sqrt{x} \\ {(5)}^{2} & = & {(\sqrt{x})}^{2} \end{array}$

x = 25

Solve. Graph the solution on a number line.

$\sqrt{4 x + 8} - 6 \leq 2$

$Restrictions Solve \begin{matrix} \sqrt{4 x + 8} \geq 0 & \sqrt{4 x + 8} \leq 8 \\ {(\sqrt{4 x + 8})}^{2} \geq {(0)}^{2} & {(\sqrt{4 x + 8})}^{2} \leq {(8)}^{2} \\ 4 x + 8 \geq 0 & 4 x + 8 \leq 64 \\ 4 x \geq - 8 & 4 x \leq 56 \\ x \geq - 2 & x \leq 14 \end{matrix}$

$\sqrt{2 x - 15} - \sqrt{9 + x} > 0$

$\begin{array}{rcl} \begin{matrix} Restrictions & Solve \\ 2 x - 15 \geq 0 & 9 + x \geq 0 & \sqrt{2 x - 15} > \sqrt{9 + x} \\ 2 x \geq 15 & x \geq - 9 & {(\sqrt{2 x - 15})}^{2} > {(\sqrt{9 + x})}^{2} \\ x \geq \frac{15}{2} & 2 x - 15 > 9 + x \\ x > 24 \end{matrix} \end{array}$

$3 + \sqrt{\frac{1}{2} x + 7} \geq 5$

$\begin{matrix} Restriction & Solve \\ \frac{1}{2} x + 7 \geq 0 & \sqrt{\frac{1}{2} x + 7} \geq 2 \\ \frac{1}{2} x \geq - 7 & {(\sqrt{\frac{1}{2} x + 7})}^{2} \geq {(2)}^{2} \\ x \geq - 14 & \frac{1}{2} x + 7 \geq 4 \\ \frac{1}{2} x \geq - 3 \\ x \geq - 6 \end{matrix}$

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

$\begin{matrix} Restriction & Solve \\ 3 x - 4 \geq 0 & - \sqrt{3 x - 4} < - 3 \\ 3 x \geq 4 & \sqrt{3 x - 4} > 3 \\ x \geq \frac{4}{3} & {(\sqrt{3 x - 4})}^{2} > {(3)}^{2} \\ 3 x - 4 > 9 \\ 3 x > 13 \\ x > \frac{13}{3} \end{matrix}$