Lynn Marecek; Andrea Honeycutt Mathis

Learning Objectives

By the end of this section, you will be able to:

Solve equations in quadratic form

Be Prepared 9.10

Before you get started, take this readiness quiz.

Factor by substitution: $y^{4} - y^{2} - 20 .$
If you missed this problem, review Example 6.21.

Be Prepared 9.11

Factor by substitution: ${(y - 4)}^{2} + 8 (y - 4) + 15 .$
If you missed this problem, review Example 6.22.

Be Prepared 9.12

Simplify: ⓐ $x^{\frac{1}{2}} \cdot x^{\frac{1}{4}}$ ⓑ ${(x^{\frac{1}{3}})}^{2}$ ⓒ ${(x^{−1})}^{2} .$
If you missed this problem, review Example 8.33.

Solve Equations in Quadratic Form

Sometimes when we factored trinomials, the trinomial did not appear to be in the ax² + bx + c form. So we factored by substitution allowing us to make it fit the ax² + bx + c form. We used the standard $u$ for the substitution.

To factor the expression x⁴ − 4x² − 5, we noticed the variable part of the middle term is x² and its square, x⁴, is the variable part of the first term. (We know ${(x^{2})}^{2} = x^{4} .$ ) So we let u = x² and factored.



Let $u = x^{2}$ and substitute.
Factor the trinomial.
Replace u with $x^{2}$ .

Similarly, sometimes an equation is not in the ax² + bx + c = 0 form but looks much like a quadratic equation. Then, we can often make a thoughtful substitution that will allow us to make it fit the ax² + bx + c = 0 form. If we can make it fit the form, we can then use all of our methods to solve quadratic equations.

Notice that in the quadratic equation ax² + bx + c = 0, the middle term has a variable, x, and its square, x², is the variable part of the first term. Look for this relationship as you try to find a substitution.

Again, we will use the standard u to make a substitution that will put the equation in quadratic form. If the substitution gives us an equation of the form ax² + bx + c = 0, we say the original equation was of quadratic form.

The next example shows the steps for solving an equation in quadratic form.

Example 9.30

How to Solve Equations in Quadratic Form

Solve: $6 x^{4} - 7 x^{2} + 2 = 0$

Solution

Step 1 is to identify a substitution that will put the equation in quadratic form. Look at the equation 6 x to the fourth power minus 7 x squared plus 2 equals 0. Since the square of x squared equals x to the fourth, let u equals x squared.

Step 2 is to rewrite the equation with the substitution to put it in quadratic form. Rewrite the equation to prepare for the substitution to show 6 times the square of x squared minus 7 times x squared plus 2 equals 0. Substitute u equals x squared to get the new equation 6 times u squared minus 7 u plus 2 equals 0.

Step 3 is to solve the quadratic equation for u. We can solve by factoring, so rewrite the equation as the product of 2 u minus 1 and 3 u minus 2 equals 0. Use the Zero Product Property to create 2 equations. If 2 u minus 1 equals 0, then 2u equals one, so u equals one half. If 3 u minus 2 equals 0, then 3 u equals 2 and u equals two thirds.

In step 4, substitute the original variable back into the results. In this case replace u with x squared. So u equals one half becomes x squared equals one half and u equals two thirds becomes x squared equals two thirds.

Step 5 is to solve for the original variable, so use the Square Root Property to solve for x. If x squared equals one half, then x equals the positive or negative square root of one half. Rationalize the denominator to see that x equals the positive or negative square root of 2 divided by 2. If x squared equals two thirds, then x equals the positive or negative square root of two thirds. Rationalize the denominator to see that x equals the positive or negative square root of 6 divided by 3.

In step 6, check your solutions. We will show one check here, x equals square root 2 divided by 2. Substitute this value into the original equation. 6 times the fourth power of the quotient square root 2 divided by 2 minus 7 times the square of the quotient square root of 2 divided by 2 plus 2. Does this expression equal 0? Simplify the powers. 6 times four sixteenths minus 7 times two fourths plus 2. Simplify terms. Three halves minus seven halves plus four halves equals zero. Square root 2 divided by 2 is a solution. We leave the other checks to you!

Try It 9.59

Solve: $x^{4} - 6 x^{2} + 8 = 0$ .

Try It 9.60

Solve: $x^{4} - 11 x^{2} + 28 = 0$ .

We summarize the steps to solve an equation in quadratic form.

How To

Solve equations in quadratic form.

Step 1. Identify a substitution that will put the equation in quadratic form.
Step 2. Rewrite the equation with the substitution to put it in quadratic form.
Step 3. Solve the quadratic equation for u.
Step 4. Substitute the original variable back into the results, using the substitution.
Step 5. Solve for the original variable.
Step 6. Check the solutions.

In the next example, the binomial in the middle term, (x − 2) is squared in the first term. If we let u = x − 2 and substitute, our trinomial will be in ax² + bx + c form.

Example 9.31

Solve: ${(x - 2)}^{2} + 7 (x - 2) + 12 = 0 .$

Solution


Prepare for the substitution.
Let $u = x - 2$ and substitute.
Solve by factoring.
Replace $u$ with $x - 2 .$
Solve for $x .$
Check:

Try It 9.61

Solve: ${(x - 5)}^{2} + 6 (x - 5) + 8 = 0 .$

Try It 9.62

Solve: ${(y - 4)}^{2} + 8 (y - 4) + 15 = 0 .$

In the next example, we notice that ${(\sqrt{x})}^{2} = x .$ Also, remember that when we square both sides of an equation, we may introduce extraneous roots. Be sure to check your answers!

Example 9.32

Solve: $x - 3 \sqrt{x} + 2 = 0 .$

Solution

The $\sqrt{x}$ in the middle term, is squared in the first term ${(\sqrt{x})}^{2} = x .$ If we let $u = \sqrt{x}$ and substitute, our trinomial will be in ax² + bx + c = 0 form.


Rewrite the trinomial to prepare for the substitution.
Let $u = \sqrt{x}$ and substitute.
Solve by factoring.
Replace u with $\sqrt{x} .$
Solve for x, by squaring both sides.
Check:

Try It 9.63

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

Try It 9.64

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

Substitutions for rational exponents can also help us solve an equation in quadratic form. Think of the properties of exponents as you begin the next example.

Example 9.33

Solve: $x^{\frac{2}{3}} - 2 x^{\frac{1}{3}} - 24 = 0 .$

Solution

The $x^{\frac{1}{3}}$ in the middle term is squared in the first term ${(x^{\frac{1}{3}})}^{2} = x^{\frac{2}{3}} .$ If we let $u = x^{\frac{1}{3}}$ and substitute, our trinomial will be in ax² + bx + c = 0 form.


Rewrite the trinomial to prepare for the substitution.
Let $u = x^{\frac{1}{3}}$ and substitute.
Solve by factoring.
Replace u with $x^{\frac{1}{3}} .$
Solve for $x$ by cubing both sides.
Check:

Try It 9.65

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

Try It 9.66

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

In the next example, we need to keep in mind the definition of a negative exponent as well as the properties of exponents.

Example 9.34

Solve: $3 x^{−2} - 7 x^{−1} + 2 = 0 .$

Solution

The $x^{−1}$ in the middle term is squared in the first term ${(x^{−1})}^{2} = x^{−2} .$ If we let $u = x^{−1}$ and substitute, our trinomial will be in ax² + bx + c = 0 form.


Rewrite the trinomial to prepare for the substitution.
Let $u = x^{−1}$ and substitute.
Solve by factoring.


Replace u with $x^{−1} .$
Solve for $x$ by taking the reciprocal since $x^{- 1} = \frac{1}{x} .$
Check:

Try It 9.67

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

Try It 9.68

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

Media

Access this online resource for additional instruction and practice with solving quadratic equations.

Solving Equations in Quadratic Form

Section 9.4 Exercises

Practice Makes Perfect

Solve Equations in Quadratic Form

In the following exercises, solve.

155.

$x^{4} - 7 x^{2} + 12 = 0$

156.

$x^{4} - 9 x^{2} + 18 = 0$

157.

$x^{4} - 13 x^{2} - 30 = 0$

158.

$x^{4} + 5 x^{2} - 36 = 0$

159.

$2 x^{4} - 5 x^{2} + 3 = 0$

160.

$4 x^{4} - 5 x^{2} + 1 = 0$

161.

$2 x^{4} - 7 x^{2} + 3 = 0$

162.

$3 x^{4} - 14 x^{2} + 8 = 0$

163.

${(x - 3)}^{2} - 5 (x - 3) - 36 = 0$

164.

${(x + 2)}^{2} - 3 (x + 2) - 54 = 0$

165.

${(3 y + 2)}^{2} + (3 y + 2) - 6 = 0$

166.

${(5 y - 1)}^{2} + 3 (5 y - 1) - 28 = 0$

167.

${(x^{2} + 1)}^{2} - 5 (x^{2} + 1) + 4 = 0$

168.

${(x^{2} - 4)}^{2} - 4 (x^{2} - 4) + 3 = 0$

169.

$2 {(x^{2} - 5)}^{2} - 5 (x^{2} - 5) + 2 = 0$

170.

$2 {(x^{2} - 5)}^{2} - 7 (x^{2} - 5) + 6 = 0$

171.

$x - \sqrt{x} - 20 = 0$

172.

$x - 8 \sqrt{x} + 15 = 0$

173.

$x + 6 \sqrt{x} - 16 = 0$

174.

$x + 4 \sqrt{x} - 21 = 0$

175.

$6 x + \sqrt{x} - 2 = 0$

176.

$6 x + \sqrt{x} - 1 = 0$

177.

$10 x - 17 \sqrt{x} + 3 = 0$

178.

$12 x + 5 \sqrt{x} - 3 = 0$

179.

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

180.

$x^{\frac{2}{3}} - 3 x^{\frac{1}{3}} = 28$

181.

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

182.

$x^{\frac{2}{3}} - 11 x^{\frac{1}{3}} + 30 = 0$

183.

$6 x^{\frac{2}{3}} - x^{\frac{1}{3}} = 12$

184.

$3 x^{\frac{2}{3}} - 10 x^{\frac{1}{3}} = 8$

185.

$8 x^{\frac{2}{3}} - 43 x^{\frac{1}{3}} + 15 = 0$

186.

$20 x^{\frac{2}{3}} - 23 x^{\frac{1}{3}} + 6 = 0$

187.

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

188.

$2 x - 7 x^{\frac{1}{2}} = 15$

189.

$6 x^{−2} + 13 x^{−1} + 5 = 0$

190.

$15 x^{−2} - 26 x^{−1} + 8 = 0$

191.

$8 x^{−2} - 2 x^{−1} - 3 = 0$

192.

$15 x^{−2} - 4 x^{−1} - 4 = 0$

Writing Exercises

193.

Explain how to recognize an equation in quadratic form.

194.

Explain the procedure for solving an equation in quadratic form.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table provides a checklist to evaluate mastery of the objectives of this section. Choose how would you respond to the statement “I can solve equations in quadratic form.” “Confidently,” “with some help,” or “No, I don’t get it.”

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

9.4 Solve Equations in Quadratic Form

Solve Equations in Quadratic Form

How to Solve Equations in Quadratic Form

Solution

Solve equations in quadratic form.

Solution

Solution

Solution

Solution

Section 9.4 Exercises

Practice Makes Perfect

Writing Exercises

Self Check