### Learning Objectives

- Solve equations in quadratic form

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.

Factor by substitution: ${\left(y-4\right)}^{2}+8\left(y-4\right)+15.$

If you missed this problem, review Example 6.22.

Simplify: ⓐ ${x}^{\frac{1}{2}}\xb7{x}^{\frac{1}{4}}$ ⓑ ${\left({x}^{\frac{1}{3}}\right)}^{2}$ ⓒ ${\left({x}^{\mathrm{-1}}\right)}^{2}.$

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 *ax*^{2} + *bx* + *c* form. So we factored by substitution allowing us to make it fit the *ax*^{2} + *bx* + *c* form. We used the standard $u$ for the substitution.

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

Let $u={x}^{2}$ and substitute. | |

Factor the trinomial. | |

Replace u with ${x}^{2}$. |

Similarly, sometimes an equation is not in the *ax*^{2} + *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 *ax*^{2} + *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 *ax*^{2} + *bx* + *c* = 0, the middle term has a variable, *x*, and its square, *x*^{2}, 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 *ax*^{2} + *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: $6{x}^{4}-7{x}^{2}+2=0$

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

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 *ax*^{2} + *bx* + *c* form.

### Example 9.31

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

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

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

In the next example, we notice that ${\left(\sqrt{x}\right)}^{2}=x.$ 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: $x-3\sqrt{x}+2=0.$

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

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

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

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: $3{x}^{\mathrm{-2}}-7{x}^{\mathrm{-1}}+2=0.$

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

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

### Media Access Additional Online Resources

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

### Section 9.4 Exercises

#### Practice Makes Perfect

**Solve Equations in Quadratic Form**

In the following exercises, solve.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$15{x}^{\mathrm{-2}}-26{x}^{\mathrm{-1}}+8=0$

$15{x}^{\mathrm{-2}}-4{x}^{\mathrm{-1}}-4=0$

#### Writing Exercises

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.

ⓑ 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?