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2.6 Other Types of Equations

Equations carrying rational exponents, radicals, or absolute value bars, solved by isolating and raising to a power, then checking for extraneous solutions.

Algebra and Trigonometry 2e, 2.6 Other Types of Equations

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45 exercises

Evaluating a Number Raised to a Rational Exponent

Evaluate 8 2 3 . 8^{\frac{2}{3}} .

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2.6 Exercise 2

Evaluate 64 − 1 3 . 64^{- \frac{1}{3}} .

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Solve the Equation Including a Variable Raised to a Rational Exponent

Solve the equation in which a variable is raised to a rational exponent: x 5 4 =32. x^{\frac{5}{4}} = 32.

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2.6 Exercise 4

Solve the equation x 3 2 =125. x^{\frac{3}{2}} = 125.

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Solving an Equation Involving Rational Exponents and Factoring

Solve 3 x 3 4 = x 1 2 . 3 x^{\frac{3}{4}} = x^{\frac{1}{2}} .

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2.6 Exercise 6

Solve: ( x+5 ) 3 2 =8. \left(x + 5\right)^{\frac{3}{2}} = 8.

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Solving a Polynomial by Factoring

Solve the polynomial by factoring: 5 x 4 =80 x 2 . 5 x^{4} = 80 x^{2} .

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2.6 Exercise 8

Solve by factoring: 12 x 4 =3 x 2 . 12 x^{4} = 3 x^{2} .

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Solving an Equation with One Radical

Solve 15−2x =x. \sqrt{15 - 2 x} = x .

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2.6 Exercise 11

Solve the radical equation: x+3 =3x−1 \sqrt{x + 3} = 3 x - 1

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Solving a Radical Equation Containing Two Radicals

Solve 2x+3 + x−2 =4. \sqrt{2 x + 3} + \sqrt{x - 2} = 4.

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2.6 Exercise 13

Solve the equation with two radicals: 3x+7 + x+2 =1. \sqrt{3 x + 7} + \sqrt{x + 2} = 1.

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Solving Absolute Value Equations

Solve the following absolute value equations:

  1. ⓐ | 6x+4 |=8 \left|6 x + 4\right| = 8
  2. ⓑ | 3x+4 |=−9 \left|3 x + 4\right| = −9
  3. ⓒ | 3x−5 |−4=6 \left|3 x - 5\right| - 4 = 6
  4. ⓓ | −5x+10 |=0 \left|−5 x + 10\right| = 0
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2.6 Exercise 15

Solve the absolute value equation: | 1−4x |+8=13. \left|1 - 4 x\right| + 8 = 13.

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Solving a Fourth-degree Equation in Quadratic Form

Solve this fourth-degree equation: 3 x 4 −2 x 2 −1=0. 3 x^{4} - 2 x^{2} - 1 = 0.

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2.6 Exercise 17

Solve using substitution: x 4 −8 x 2 −9=0. x^{4} - 8 x^{2} - 9 = 0.

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Solving an Equation in Quadratic Form Containing a Binomial

Solve the equation in quadratic form: ( x+2 ) 2 +11( x+2 )−12=0. \left(x + 2\right)^{2} + 11 \left(x + 2\right) - 12 = 0.

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2.6 Exercise 19

Solve: ( x−5 ) 2 −4( x−5 )−21=0. \left(x - 5\right)^{2} - 4 \left(x - 5\right) - 21 = 0.

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Solving a Rational Equation Leading to a Quadratic

Solve the following rational equation: −4x x−1 + 4 x+1 = −8 x 2 −1 . \frac{- 4 x}{x - 1} + \frac{4}{x + 1} = \frac{- 8}{x^{2} - 1} .

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2.6 Exercise 21

Solve 3x+2 x−2 + 1 x = −2 x 2 −2x . \frac{3 x + 2}{x - 2} + \frac{1}{x} = \frac{- 2}{x^{2} - 2 x} .

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2.6 Exercise 22

In a radical equation, what does it mean if a number is an extraneous solution?

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2.6 Exercise 24

Your friend tries to calculate the value − 9 3 2 - 9^{\frac{3}{2}} and keeps getting an ERROR message. What mistake are they probably making?

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2.6 Exercise 26

Explain how to change a rational exponent into the correct radical expression.

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2.6 Exercise 28

Solve the rational exponent equation. Use factoring where necessary.

x 3 4 =27 x^{\frac{3}{4}} = 27

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2.6 Exercise 30

Solve the rational exponent equation. Use factoring where necessary.

( x−1 ) 3 4 =8 \left(x - 1\right)^{\frac{3}{4}} = 8

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2.6 Exercise 32

Solve the rational exponent equation. Use factoring where necessary.

x 2 3 −5 x 1 3 +6=0 x^{\frac{2}{3}} - 5 x^{\frac{1}{3}} + 6 = 0

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2.6 Exercise 34

Solve the following polynomial equations by grouping and factoring.

x 3 +2 x 2 −x−2=0 x^{3} + 2 x^{2} - x - 2 = 0

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2.6 Exercise 36

Solve the following polynomial equations by grouping and factoring.

4 y 3 −9y=0 4 y^{3} - 9 y = 0

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2.6 Exercise 38

Solve the following polynomial equations by grouping and factoring.

m 3 + m 2 −m−1=0 m^{3} + m^{2} - m - 1 = 0

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2.6 Exercise 40

Solve the following polynomial equations by grouping and factoring.

5 x 3 +45x=2 x 2 +18 5 x^{3} + 45 x = 2 x^{2} + 18

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2.6 Exercise 42

Solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions.

x−7 =5 \sqrt{x - 7} = 5

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2.6 Exercise 44

Solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions.

3t+5 =7 \sqrt{3 t + 5} = 7

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2.6 Exercise 46

Solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions.

12−x =x \sqrt{12 - x} = x

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2.6 Exercise 48

Solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions.

3x+7 + x+2 =1 \sqrt{3 x + 7} + \sqrt{x + 2} = 1

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2.6 Exercise 50

Solve the equation involving absolute value.

| 3x−4 |=8 \left|3 x - 4\right| = 8

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2.6 Exercise 52

Solve the equation involving absolute value.

| 1−4x |−1=5 \left|1 - 4 x\right| - 1 = 5

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2.6 Exercise 54

Solve the equation involving absolute value.

| 2x−1 |−7=−2 \left|2 x - 1\right| - 7 = −2

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2.6 Exercise 56

Solve the equation involving absolute value.

| x+5 |=0 \left|x + 5\right| = 0

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2.6 Exercise 58

Solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring.

x 4 −10 x 2 +9=0 x^{4} - 10 x^{2} + 9 = 0

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2.6 Exercise 60

Solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring.

( x 2 −1 ) 2 +( x 2 −1 )−12=0 \left(x^{2} - 1\right)^{2} + \left(x^{2} - 1\right) - 12 = 0

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2.6 Exercise 62

Solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring.

( x−3 ) 2 −4=0 \left(x - 3\right)^{2} - 4 = 0

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2.6 Exercise 64

Solve for the unknown variable.

| x | 2 =x \sqrt{\left|x\right|^{2}} = x

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2.6 Exercise 66

Solve for the unknown variable.

| x 2 +2x−36 |=12 \left|x^{2} + 2 x - 36\right| = 12

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2.6 Exercise 68

Use the model for the period of a pendulum, T, T , such that T=2π L g , T = 2 \pi \sqrt{\frac{L}{g}} , where the length of the pendulum is L and the acceleration due to gravity is g. g .

If the gravity is 32 ft/s2 and the period equals 1 s, find the length to the nearest in. (12 in. = 1 ft). Round your answer to the nearest in.

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2.6 Exercise 70

Use a model for body surface area, BSA, such that BSA= wh 3600 , B S A = \sqrt{\frac{w h}{3600}} , where w = weight in kg and h = height in cm.

Find the weight of a 177-cm male to the nearest kg whose BSA=2.1. B S A = 2.1.

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Adapted from College Algebra © 2021 Rice University. Licensed under CC BY 4.0. Sources: College Algebra 2e, Algebra and Trigonometry 2e, Precalculus 2e, College Algebra with Corequisite Support 2e. Imported at revision 4922e46.