9.4 Solve Equations in Quadratic Form

Learning objectives

Learning objectives

• Solve equations in quadratic form

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 ${({x}^{2})}^{2}={x}^{4}.$) So we let u = x${}^{2}$ and factored.

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

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

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

Solution

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

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 ax${}^{2}$ + bx + c = 0 form.

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

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 ax${}^{2}$ + bx + c = 0 form.

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

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 ax${}^{2}$ + bx + c = 0 form.

Section 9.4 Exercises

Section 9.4 Exercises

Solve Equations in Quadratic Form

In the following exercises, solve.

Writing Exercises

ⓐ 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?

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

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

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

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

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

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

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

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

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

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

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

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

1. $2{x}^{4}-5{x}^{2}+3=0$

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

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

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

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

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

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

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

1. $6x+\sqrt{x}-2=0$

2. $10x-17\sqrt{x}+3=0$

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

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

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

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

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

2. $6{x}^{-2}+13{x}^{-1}+5=0$

1. $8{x}^{-2}-2{x}^{-1}-3=0$

2. Explain how to recognize an equation in quadratic form.