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9.4 Solve Equations in Quadratic Form

Substituting a single variable for a repeated expression so fourth-degree, radical, and rational-exponent equations collapse into a quadratic.

Intermediate Algebra 2e, 9.4 Solve Equations in Quadratic Form

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33 exercises

9.4 Exercise 1

Before you get started, take this readiness quiz.

Factor by substitution: y4−y2−20.y^{4} - y^{2} - 20 .
If you missed this problem, review the reference.

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9.4 Exercise 2

Factor by substitution: (y−4)2+8(y−4)+15.\left(y - 4\right)^{2} + 8 \left(y - 4\right) + 15 .
If you missed this problem, review the reference.

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9.4 Exercise 3

Simplify: ⓐ x12·x14x^{\frac{1}{2}} \cdot x^{\frac{1}{4}} ⓑ (x13)2\left(x^{\frac{1}{3}}\right)^{2} ⓒ (x−1)2.\left(x^{−1}\right)^{2} .
If you missed this problem, review the reference.

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9.4 Exercise 5

Solve: x4−6x2+8=0x^{4} - 6 x^{2} + 8 = 0.

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9.4 Exercise 6

Solve: x4−11x2+28=0x^{4} - 11 x^{2} + 28 = 0.

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9.4 Exercise 8

Solve: (x−5)2+6(x−5)+8=0.\left(x - 5\right)^{2} + 6 \left(x - 5\right) + 8 = 0 .

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9.4 Exercise 9

Solve: (y−4)2+8(y−4)+15=0.\left(y - 4\right)^{2} + 8 \left(y - 4\right) + 15 = 0 .

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9.4 Exercise 11

Solve: x−7x+12=0.x - 7 \sqrt{x} + 12 = 0 .

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9.4 Exercise 12

Solve: x−6x+8=0.x - 6 \sqrt{x} + 8 = 0 .

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9.4 Exercise 14

Solve: x23−5x13−14=0.x^{\frac{2}{3}} - 5 x^{\frac{1}{3}} - 14 = 0 .

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9.4 Exercise 15

Solve: x12−8x14+15=0.x^{\frac{1}{2}} - 8 x^{\frac{1}{4}} + 15 = 0 .

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9.4 Exercise 17

Solve: 8x−2−10x−1+3=0.8 x^{−2} - 10 x^{−1} + 3 = 0 .

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9.4 Exercise 18

Solve: 6x−2−23x−1+20=0.6 x^{−2} - 23 x^{−1} + 20 = 0 .

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9.4 Exercise 19

Solve.

x4−7x2+12=0x^{4} - 7 x^{2} + 12 = 0

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9.4 Exercise 21

Solve.

x4−13x2−30=0x^{4} - 13 x^{2} - 30 = 0

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9.4 Exercise 23

Solve.

2x4−5x2+3=02 x^{4} - 5 x^{2} + 3 = 0

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9.4 Exercise 25

Solve.

2x4−7x2+3=02 x^{4} - 7 x^{2} + 3 = 0

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9.4 Exercise 27

Solve.

(x−3)2−5(x−3)−36=0\left(x - 3\right)^{2} - 5 \left(x - 3\right) - 36 = 0

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9.4 Exercise 29

Solve.

(3y+2)2+(3y+2)−6=0\left(3 y + 2\right)^{2} + \left(3 y + 2\right) - 6 = 0

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9.4 Exercise 31

Solve.

(x2+1)2−5(x2+1)+4=0\left(x^{2} + 1\right)^{2} - 5 \left(x^{2} + 1\right) + 4 = 0

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9.4 Exercise 33

Solve.

2(x2−5)2−5(x2−5)+2=02 \left(x^{2} - 5\right)^{2} - 5 \left(x^{2} - 5\right) + 2 = 0

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9.4 Exercise 35

Solve.

x−x−20=0x - \sqrt{x} - 20 = 0

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9.4 Exercise 37

Solve.

x+6x−16=0x + 6 \sqrt{x} - 16 = 0

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9.4 Exercise 39

Solve.

6x+x−2=06 x + \sqrt{x} - 2 = 0

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9.4 Exercise 41

Solve.

10x−17x+3=010 x - 17 \sqrt{x} + 3 = 0

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9.4 Exercise 43

Solve.

x23+9x13+8=0x^{\frac{2}{3}} + 9 x^{\frac{1}{3}} + 8 = 0

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9.4 Exercise 45

Solve.

x23+4x13=12x^{\frac{2}{3}} + 4 x^{\frac{1}{3}} = 12

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9.4 Exercise 47

Solve.

6x23−x13=126 x^{\frac{2}{3}} - x^{\frac{1}{3}} = 12

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9.4 Exercise 49

Solve.

8x23−43x13+15=08 x^{\frac{2}{3}} - 43 x^{\frac{1}{3}} + 15 = 0

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9.4 Exercise 51

Solve.

x−8x12+7=0x - 8 x^{\frac{1}{2}} + 7 = 0

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9.4 Exercise 53

Solve.

6x−2+13x−1+5=06 x^{−2} + 13 x^{−1} + 5 = 0

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9.4 Exercise 55

Solve.

8x−2−2x−1−3=08 x^{−2} - 2 x^{−1} - 3 = 0

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9.4 Exercise 57

Explain how to recognize an equation in quadratic form.

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Adapted from Intermediate Algebra 2e © 2020 Rice University. Licensed under CC BY 4.0. Source: Intermediate Algebra 2e. Imported at revision 77a933c.