Practice 2 Solutions

Explain why the absolute value of x must be taken when n is an even exponent.
Sample: When the base is raised to an even exponent, it must be positive.

Solve. Check for extraneous solutions.

${(4 x)}^{\frac{1}{3}} = {(3 x - 4)}^{\frac{1}{3}}$

$\begin{array}{rcl} \begin{matrix} {({(4 x)}^{\frac{1}{3}})}^{3} = {({(3 x - 4)}^{\frac{1}{3}})}^{3} & Check \\ 4 x = 3 x - 4 & {(4 (- 4))}^{\frac{1}{3}} = {(3 (- 4) - 4)}^{\frac{1}{3}} \\ x = - 4 & {(- 16)}^{\frac{1}{3}} = {(- 16)}^{\frac{1}{3}} ✓ \end{matrix} \end{array}$

x = – 4

Note

Remember that it is possible to take an odd root of a negative number because an odd number of negative numbers will always be negative.

$\sqrt{h^{2} - 9} + 3 = h$

$\begin{array}{rcl} \begin{matrix} \sqrt{h^{2} - 9} = h - 3 & Check \\ {(\sqrt{h^{2} - 9})}^{2} = {(h - 3)}^{2} & \sqrt{{(3)}^{2} - 9} + 3 = 3 \\ h^{2} - 9 = h^{2} - 6 h + 9 & \sqrt{9 - 9} + 3 = 3 \\ 6 h = 18 & \sqrt{0} + 3 = 3 ✓ \\ h = 3 \end{matrix} \end{array}$

h = 3

$\sqrt{a} = \sqrt{a - 8} - 2$

$\begin{array}{rcl} \sqrt{a} + 2 & = & \sqrt{a - 8} \\ {(\sqrt{a} + 2)}^{2} & = & {(\sqrt{a - 8})}^{2} \\ a + 4 \sqrt{a} + 4 & = & a - 8 \\ 4 \sqrt{a} & = & - 12 \\ \sqrt{a} & = & - 3 \end{array}$

No real solution

Note

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

${(2 w - 5)}^{\frac{2}{3}} = 9$

${({(2 w - 5)}^{\frac{2}{3}})}^{\frac{3}{2}} = 9^{\frac{3}{2}} 9^{\frac{3}{2}} = {(9^{\frac{1}{2}})}^{3} = 3^{3} = 27 |2 w - 5| = 27 \begin{matrix} case 1 & case 2 \\ 2 w - 5 = 27 & - (2 w - 5) = 27 \\ 2 w - 5 = - 27 \\ 2 w = 32 & 2 w = - 22 \\ w = 16 & w = - 11 \end{matrix}$

$\begin{array}{rcl} Check \\ {(2 (- 11) - 5)}^{\frac{2}{3}} & = & 9 \\ {(- 22 - 5)}^{\frac{2}{3}} & = & 9 \\ {(- 27)}^{\frac{2}{3}} & = & 9 \\ {(\sqrt[3]{- 27})}^{2} & = & 9 \\ {(- 3)}^{2} & = & 9 ✓ \\ {(2 (16) - 5)}^{\frac{2}{3}} & = & 9 \\ {(32 - 5)}^{\frac{2}{3}} & = & 9 \\ {(27)}^{\frac{2}{3}} & = & 9 \\ {(\sqrt[3]{27})}^{2} & = & 9 \\ 3^{2} & = & 9 ✓ \end{array}$

w = – 11, 16

Note

Problems 6–12

Recall that ${(x^{\frac{n}{d}})}^{\frac{d}{n}} = |x|$ , when n is an even number. A negative number raised to an even exponent is positive.

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

$\begin{array}{rcl} \begin{matrix} \sqrt{x} - 1 = \sqrt{x - 4} & Check \\ {(\sqrt{x} - 1)}^{2} = {(\sqrt{x - 4})}^{2} & \sqrt{\frac{25}{4}} - \sqrt{\frac{25}{4} - 4} = 1 \\ x - 2 \sqrt{x} + 1 = x - 4 & \frac{5}{2} - \sqrt{\frac{25}{4} - \frac{16}{4}} = 1 \\ - 2 \sqrt{x} = - 5 & \frac{5}{2} - \sqrt{\frac{9}{4}} = 1 \\ \sqrt{x} = \frac{5}{2} & \frac{5}{2} - \frac{3}{2} = 1 \\ {(\sqrt{x})}^{2} = {(\frac{5}{2})}^{2} & \frac{2}{2} = 1 ✓ \\ x = \frac{25}{4} \end{matrix} \end{array}$

$x = \frac{25}{4}$

${(m + 7)}^{\frac{1}{2}} - 5 = m$

$\begin{array}{rcl} \begin{matrix} {(m + 7)}^{\frac{1}{2}} = m + 5 & Check \\ {({(m + 7)}^{\frac{1}{2}})}^{2} = {(m + 5)}^{2} & {(- 6 + 7)}^{\frac{1}{2}} - 5 = - 6 \\ m + 7 = m^{2} + 10 m + 25 & {(1)}^{\frac{1}{2}} - 5 = - 6 \\ 0 = m^{2} + 9 m + 18 & 1 - 5 \neq - 6 ✖ \\ 0 = (m + 6) (m + 3) \\ m = - 6, - 3 & {(- 3 + 7)}^{\frac{1}{2}} - 5 = - 3 {(4)}^{\frac{1}{2}} - 5 = - 3 2 - 5 = - 3 ✓ \end{matrix} \end{array}$

m = – 3

$2 {(x - 2)}^{\frac{3}{2}} = 54$

$\begin{matrix} {(x - 2)}^{\frac{3}{2}} = 27 & Check \\ {({(x - 2)}^{\frac{3}{2}})}^{\frac{2}{3}} = 27^{\frac{2}{3}} & \begin{array}{rcl} 2 {(11 - 2)}^{\frac{3}{2}} & = & 54 \end{array} \\ 27^{\frac{2}{3}} = {(27^{\frac{1}{3}})}^{2} = 3^{2} = 9 & \begin{array}{rcl} 2 {(9)}^{\frac{3}{2}} & = & 54 \end{array} \\ x - 2 = 9 & \begin{array}{rcl} 2 {(\sqrt{9})}^{3} & = & 54 \end{array} \\ x = 11 & \begin{array}{rcl} 2 {(3)}^{3} & = & 54 \end{array} \\ \begin{array}{rcl} 2 (27) & = & 54 ✓ \end{array} \end{matrix}$

x = 11

${(p + 3)}^{\frac{1}{4}} + 5 = 3$

${(p + 3)}^{\frac{1}{4}} = - 2$

No solution. An even root cannot equal a negative number.

$\sqrt{4 x + 8} = 1$

$\begin{array}{rcl} \begin{matrix} {(\sqrt{4 x + 8})}^{2} = 1^{2} & Check \\ 4 x + 8 = 1 & \sqrt{4 (- \frac{7}{4}) + 8} = 1 \\ 4 x = - 7 & \sqrt{- 7 + 8} = 1 \\ x = - \frac{7}{4} & \sqrt{1} = 1 ✓ \end{matrix} \end{array}$

$x = - \frac{7}{4}$

$6 - \sqrt{3 u + 1} = 1$

$\begin{array}{rcl} \begin{matrix} - \sqrt{3 u + 1} = - 5 & Check \\ \sqrt{3 u + 1} = 5 & 6 - \sqrt{3 (8) + 1} = 1 \\ {(\sqrt{3 u + 1})}^{2} = 5^{2} & 6 - \sqrt{24 + 1} = 1 \\ 3 u + 1 = 25 & 6 - \sqrt{25} = 1 \\ 3 u = 24 & 6 - 5 = 1 ✓ \\ u = 8 \end{matrix} \end{array}$

u = 8

${(b - 3)}^{\frac{4}{5}} - 3 = 13$

${(b - 3)}^{\frac{4}{5}} = 16 {({(b - 3)}^{\frac{4}{5}})}^{\frac{5}{4}} = 16^{\frac{5}{4}} 16^{\frac{5}{4}} = {(16^{\frac{1}{4}})}^{5} = 2^{5} = 32 |b - 3| = 32 \begin{matrix} case 1 & case 2 \\ b - 3 = 32 & - (b - 3) = 32 \\ b - 3 = - 32 \\ b = 35 & b = - 29 \end{matrix}$

$\begin{array}{rcl} Check \\ {(35 - 3)}^{\frac{4}{5}} - 3 & = & 13 \\ {(32)}^{\frac{4}{5}} - 3 & = & 13 \\ {(\sqrt[5]{32})}^{4} - 3 & = & 13 \\ {(2)}^{4} - 3 & = & 13 \\ 16 - 3 & = & 13 ✓ \\ {(- 29 - 3)}^{\frac{4}{5}} - 3 & = & 13 \\ {(- 32)}^{\frac{4}{5}} - 3 & = & 13 \\ {(\sqrt[5]{- 32})}^{4} - 3 & = & 13 \\ {(- 2)}^{4} - 3 & = & 13 \\ 16 - 3 & = & 13 \end{array}$

b = – 29, 35

$\sqrt{y + 2} = \sqrt{2 y - 3}$

$\begin{array}{l} \begin{matrix} {(\sqrt{y + 2})}^{2} = {(\sqrt{2 y - 3})}^{2} & Check \\ y + 2 = 2 y - 3 & \sqrt{5 + 2} = \sqrt{2 (5) - 3} \\ y = 5 & \sqrt{7} = \sqrt{7} ✓ \end{matrix} \end{array}$

y = 5

$\sqrt{11 q + 3} = 2 q$

$\begin{array}{rcl} \begin{matrix} {(\sqrt{11 q + 3})}^{2} = {(2 q)}^{2} & Check \\ 11 q + 3 = 4 q^{2} & \sqrt{11 (- \frac{1}{4}) + 3} = 2 (- \frac{1}{4}) \\ 0 = 4 q^{2} - 11 q - 3 & \sqrt{- \frac{11}{4} + \frac{12}{4}} = - \frac{1}{2} \\ (4 q + 1) (q - 3) = 0 & \sqrt{\frac{1}{4}} = - \frac{1}{2} \\ q = - \frac{1}{4}, 3 & \frac{1}{2} \neq - \frac{1}{2} ✖ \\ \sqrt{11 (3) + 3} = 2 (3) \sqrt{33 + 3} = 6 \sqrt{36} = 6 ✓ \end{matrix} \end{array}$

$q = 3$

$\sqrt{5 + x} = \sqrt{x} + 1$

$\begin{array}{rcl} \begin{matrix} {(\sqrt{5 + x})}^{2} = {(\sqrt{x} + 1)}^{2} & Check \\ 5 + x = x + 2 \sqrt{x} + 1 & \sqrt{5 + 4} = \sqrt{4} + 1 \\ 4 = 2 \sqrt{x} & \sqrt{9} = 2 + 1 \\ 2 = \sqrt{x} & 3 = 3 ✓ \\ 2^{2} = {(\sqrt{x})}^{2} \\ x = 4 \end{matrix} \end{array}$

x = 4

The area of the circular base of a cone is $25 π {in}^{2}$ . The height of the cone is 2 times the radius value. Use the formula $A = π {(\frac{3 V}{2 π})}^{\frac{2}{3}}$ to find the volume of the cone. Keep your answer in terms of $π .$

$25 π = π {(\frac{3 V}{2 π})}^{\frac{2}{3}} 25 = {(\frac{3 V}{2 π})}^{\frac{2}{3}} {(25)}^{\frac{3}{2}} = {({(\frac{3 V}{2 π})}^{\frac{2}{3}})}^{\frac{3}{2}} 125 = \frac{3 V}{2 π} V = \frac{250 π}{3}$

$V = \frac{250 π}{3} {in}^{3}$

Note

Q: Why was it not necessary to determine Case 1 and Case 2?

A: Case 2 would be extraneous. Volume cannot be negative.