Practice 1 Solutions

Explain how radicals and rational exponents are related.

Sample: Radicals can be rewritten using the index. The index is the denominator of a fractional exponent.
$\sqrt[d]{x^{n}} = x^{\frac{n}{d}}$

Solve. Check for extraneous solutions.

Note

Remember to check solutions using mental math, a calculator, or writing out the steps using substitution. You must perform a check before you can determine a final answer.

$\sqrt{3 x + 1} - 4 = 2$

$\begin{array}{l} \begin{matrix} \sqrt{3 x + 1} = 6 & Check \\ {(\sqrt{3 x + 1})}^{2} = 6^{2} & \sqrt{3 (\frac{35}{3}) + 1} - 4 = 2 \\ 3 x + 1 = 36 & \sqrt{35 + 1} - 4 = 2 \\ 3 x = 35 & \sqrt{36} - 4 = 2 \\ x = \frac{35}{3} & 6 - 4 = 2 ✓ \end{matrix} \end{array}$

$x = \frac{35}{3}$

Note

Q: What is your first step (before squaring both sides)?

A: Adding 4 to both sides to isolate the radical.

$5 \sqrt{2 y - 8} = 15$

$\begin{array}{rcl} \begin{matrix} \sqrt{2 y - 8} = 3 & Check \\ {(\sqrt{2 y - 8})}^{2} = 3^{2} & 5 \sqrt{2 (\frac{17}{2}) - 8} = 15 \\ 2 y - 8 = 9 & 5 \sqrt{17 - 8} = 15 \\ 2 y = 17 & 5 \sqrt{9} = 15 \\ y = \frac{17}{2} & 5 (3) = 15 ✓ \end{matrix} \end{array}$

$y = \frac{17}{2}$

${(x + 6)}^{\frac{1}{2}} = x + 4$

$\begin{array}{rcl} \begin{matrix} {({(x + 6)}^{\frac{1}{2}})}^{2} = {(x + 4)}^{2} & Check \\ x + 6 = x^{2} + 8 x + 16 & {(- 5 + 6)}^{\frac{1}{2}} = - 5 + 4 \\ 0 = x^{2} + 7 x + 10 & {(1)}^{\frac{1}{2}} \neq - 1 ✖ \\ 0 = (x + 5) (x + 2) \\ x = - 5, - 2 & {(- 2 + 6)}^{\frac{1}{2}} = - 2 + 4 \\ {(4)}^{\frac{1}{2}} = 2 ✓ \end{matrix} \end{array}$

x = – 2

Note

Q: Why is it important to check your answers when working with radicals or fractional exponents?

A: Because it is possible to have extraneous solutions.

Note

Problems 5–6

Remember that the result of a binomial squared is a trinomial. Then combine like terms to finish solving for the variable.

$\sqrt{2 n - 6} + n = 3$

$\begin{array}{rcl} \begin{matrix} \sqrt{2 n - 6} = 3 - n & Check \\ {(\sqrt{2 n - 6})}^{2} = {(3 - n)}^{2} & \sqrt{2 (3) - 6} + 3 = 3 \\ 2 n - 6 = 9 - 6 n + n^{2} & \sqrt{6 - 6} + 3 = 3 \\ 0 = n^{2} - 8 n + 15 & \sqrt{0} + 3 = 3 ✓ \\ 0 = (n - 3) (n - 5) \\ n = 3, 5 & \sqrt{2 (5) - 6} + 5 = 3 \\ \sqrt{10 - 6} + 5 = 3 \\ \sqrt{4} + 5 = 3 \\ 2 + 5 \neq 3 ✖ \end{matrix} \end{array}$

n = 3

$\sqrt{2 z + 4} - \sqrt{z} = 2$

$\begin{array}{rcl} \begin{matrix} \sqrt{2 z + 4} = 2 + \sqrt{z} & Check \\ {(\sqrt{2 z + 4})}^{2} = {(2 + \sqrt{z})}^{2} & \sqrt{2 (0) + 4} - \sqrt{0} = 2 \\ 2 z + 4 = 4 + 4 \sqrt{z} + z & \sqrt{4} = 2 ✓ \\ z = 4 \sqrt{z} \\ {(z)}^{2} = {(4 \sqrt{z})}^{2} & \sqrt{2 (16) + 4} - \sqrt{16} = 2 \\ z^{2} = 16 z & \sqrt{32 + 4} - 4 = 2 \\ z^{2} - 16 z = 0 & \sqrt{36} - 4 = 2 \\ z (z - 16) = 0 & 6 - 4 = 2 ✓ \\ z = 0, z = 16 \end{matrix} \end{array}$

z = 0, 16

$8 {(3 x - 1)}^{\frac{1}{5}} + 2 = 18$

$\begin{array}{rcl} \begin{matrix} 8 {(3 x - 1)}^{\frac{1}{5}} = 16 & Check \\ {(3 x - 1)}^{\frac{1}{5}} = 2 & 8 {(3 (11) - 1)}^{\frac{1}{5}} + 2 = 18 \\ {({(3 x - 1)}^{\frac{1}{5}})}^{5} = 2^{5} & 8 {(33 - 1)}^{\frac{1}{5}} + 2 = 18 \\ 3 x - 1 = 32 & 8 {(32)}^{\frac{1}{5}} + 2 = 18 \\ 3 x = 33 & 8 (2) + 2 = 18 \\ x = 11 & 16 + 2 = 18 ✓ \end{matrix} \end{array}$

x = 11

Note

Q: Why is 2 raised to the 5th power not the same as 2 multiplied by 5?

A: Sample: Because 2⁵ is the same as $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ which is 32. This is different from 2 fives, or $2 \cdot 5 = 10$ .

$\sqrt{y + 7} + 5 = y$

$\begin{array}{rcl} \begin{matrix} \sqrt{y + 7} = y - 5 & Check \\ {(\sqrt{y + 7})}^{2} = {(y - 5)}^{2} & \sqrt{2 + 7} + 5 = 2 \\ y + 7 = y^{2} - 10 y + 25 & \sqrt{9} + 5 = 2 \\ 0 = y^{2} - 11 y + 18 & 3 + 5 \neq 2 ✖ \\ 0 = (y - 2) (y - 9) \\ y = 2, 9 & \sqrt{9 + 7} + 5 = 9 \\ \sqrt{16} + 5 = 9 \\ 4 + 5 = 9 ✓ \end{matrix} \end{array}$

y = 9

$\sqrt{4 x + 5} = x + 2$

$\begin{array}{rcl} \begin{matrix} {(\sqrt{4 x + 5})}^{2} = {(x + 2)}^{2} & Check \\ 4 x + 5 = x^{2} + 4 x + 4 & \sqrt{4 (- 1) + 5} = - 1 + 2 \\ 0 = x^{2} - 1 & \sqrt{- 4 + 5} = 1 \\ 0 = (x + 1) (x - 1) & \sqrt{1} = 1 ✓ \\ x = \pm 1 \\ \sqrt{4 (1) + 5} = 1 + 2 \sqrt{4 + 5} = 1 + 2 \sqrt{9} = 3 ✓ \end{matrix} \end{array}$

x = ± 1

$2 {(r + 3)}^{\frac{3}{2}} = 54$

$\begin{array}{rcl} \begin{matrix} {(r + 3)}^{\frac{3}{2}} = 27 & Check \\ {({(r + 3)}^{\frac{3}{2}})}^{\frac{2}{3}} = 27^{\frac{2}{3}} & 2 {(6 + 3)}^{\frac{3}{2}} = 54 \\ 27^{\frac{2}{3}} = {(27^{\frac{1}{3}})}^{2} = 3^{2} = 9 & 2 {(9)}^{\frac{3}{2}} = 54 \\ r + 3 = 9 & 2 (27) = 54 ✓ \\ r = 6 \end{matrix} \end{array}$

r = 6

Note

Q: What is the reciprocal of the exponent?

A: two-thirds

It may be helpful to write $27^{\frac{2}{3}} = {(\sqrt[3]{27})}^{2} = 3^{2} = 9$ so you can simplify this number completely before solving for r. When checking, it may be helpful to write ${(9)}^{\frac{3}{2}} = {(\sqrt{9})}^{3} = 3^{3} = 27$ .

${(6 x - 5)}^{\frac{1}{2}} + 3 = - 2$

${(6 x - 5)}^{\frac{1}{2}} = - 5$

No real solution

Note

A principal (positive) square root cannot equal a negative number. If you solved the problem and found that x = 5, you will see when checking that this is not possible.

$\sqrt{6 x - 3} = \sqrt{5 x + 2}$

$\begin{array}{rcl} \begin{matrix} {(\sqrt{6 x - 3})}^{2} = {(\sqrt{5 x + 2})}^{2} & Check \\ 6 x - 3 = 5 x + 2 & \sqrt{6 (5) - 3} = \sqrt{5 (5) + 2} \\ x = 5 & \sqrt{27} = \sqrt{27} ✓ \end{matrix} \end{array}$

x = 5

$5 {(w - 3)}^{\frac{2}{3}} = 45$

$\begin{array}{rcl} {(w - 3)}^{\frac{2}{3}} & = & 9 \\ {({(w - 3)}^{\frac{2}{3}})}^{\frac{3}{2}} & = & 9^{\frac{3}{2}} \\ 9^{\frac{3}{2}} & = & {(9^{\frac{1}{2}})}^{3} = 3^{3} = 27 \\ | w - 3 | & = & 27 \\ \begin{matrix} case 1 \\ w - 3 = 27 \\ w = 30 \end{matrix} \begin{matrix} case 2 \\ - (w - 3) = 27 \\ w - 3 = - 27 \\ w = - 24 \end{matrix} \end{array}$

$\begin{array}{rcl} Check \\ 5 {(30 - 3)}^{\frac{2}{3}} & = & 45 \\ 5 {(27)}^{\frac{2}{3}} & = & 45 \\ 5 {(\sqrt[3]{27})}^{2} & = & 45 \\ 5 {(3)}^{2} & = & 45 \\ 5 (9) & = & 45 ✓ \\ 5 {(- 24 - 3)}^{\frac{2}{3}} & = & 45 \\ 5 {(- 27)}^{\frac{2}{3}} & = & 45 \\ 5 {(\sqrt[3]{27})}^{2} & = & 45 \\ 5 {(- 3)}^{2} & = & 45 \\ 5 (9) & = & 45 ✓ \end{array}$

x = – 24, 30

Note

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. It is possible for cubed roots to have negative answers because a negative number to an odd power is always negative.

Q: What is the power of the fractional exponent? What does this tell you about the solution?

A: 2, this means that there are two cases because the absolute value is taken.

$2 \sqrt{a + 5} = \sqrt{3 a + 1}$

$\begin{array}{rcl} \begin{matrix} {(2 \sqrt{a + 5})}^{2} = {(\sqrt{3 a + 1})}^{2} & Check \\ 4 (a + 5) = 3 a + 1 & 2 \sqrt{- 19 + 5} = \sqrt{3 (- 19) + 1} \\ 4 a + 20 = 3 a + 1 & 2 \sqrt{- 14} = \sqrt{- 56} \\ a = - 19 \end{matrix} \end{array}$

No real solution

Note

No real solution. The square root of a negative number is not real.

The word “real” is critical here because you will learn how to solve this type of problem using complex numbers in a future lesson.

${(n - 2)}^{\frac{4}{5}} = 16$

{({(n - 2)}^{\frac{4}{5}})}^{\frac{5}{4}} = 16^{\frac{5}{4}} 16^{\frac{5}{4}} = {(16^{\frac{1}{4}})}^{5} = 2^{5} = 32 | n - 2 | = 32 \begin{matrix} case 1 & case 2 \\ n - 2 = 32 & - (n - 2) = 32 \\ n - 2 = - 32 \\ n = 34 & n = - 30 \end{matrix}

\begin{array}{rcl} Check \\ {(34 - 2)}^{\frac{4}{5}} & = & 16 \\ {(32)}^{\frac{4}{5}} & = & 16 \\ {(\sqrt[5]{32})}^{4} & = & 16 \\ 2^{4} & = & 16 ✓ \\ {(- 30 - 2)}^{\frac{4}{5}} & = & 16 \\ {(- 32)}^{\frac{4}{5}} & = & 16 \\ {(\sqrt[5]{- 32})}^{4} & = & 16 \\ {(- 2)}^{4} & = & 16 ✓ \end{array}

n = – 30, 34

Note

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.

The compound interest formula is $F = P {(1 + r)}^{n}$ , where F is the future value, P is the present value, r is the percent interest rate, and n is the number of years. What percent interest rate would you need to end with a future value of $2,151.86, when you get compound interest on $2,000 for 18 months? Round to the nearest tenth of a percent.

$18 months = \frac{3}{2} years F = 2151.86, P = 2000, n = \frac{3}{2} 2151.86 = 2000 {(1 + r)}^{\frac{3}{2}} \frac{2151.86}{2000} = {(1 + r)}^{\frac{3}{2}} {(\frac{2151.86}{2000})}^{\frac{2}{3}} = {({(1 + r)}^{\frac{3}{2}})}^{\frac{2}{3}} 1.025 = 1 + r r = 0.025$

Note

Q: Why was 0.025 converted to 25%?

A: The question asks for the percent and the rate is a decimal.

Q: Why did the problem not require Case 1 and Case 2?

A: Case 2 would have produced a negative interest rate, which would be extraneous.

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