9.4 Solve Equations in Quadratic Form

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.

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 ax² + bx + c form.

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 3

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

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:

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 4

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

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:

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

Example 5

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

The $x to the power −1$ 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 $x to the power −1.$		$Two mathematical equations are displayed: x^-1 = 1/3, and x^-1 = 2. Both equations show an inverse power of x set equal to a constant value, with the first constant being a fraction and the second an integer.$
Solve for $x$ by taking the reciprocal since $x^{- 1} = \frac{1}{x} .$
Check:

Key Concepts

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

Practice Makes Perfect

Solve Equations in Quadratic Form

In the following exercises, solve.

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

$x = \pm \sqrt{3}, x = \pm 2$

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

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

$x = \pm \sqrt{15}, x = \pm \sqrt{2} i$

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

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

$x = \pm 1, x = \frac{\pm \sqrt{6}}{2}$

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

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

$x = \pm \sqrt{3}, x = \pm \frac{\sqrt{2}}{2}$

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

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

$x = −1, x = 12$

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

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

$x = - \frac{5}{3}, x = 0$

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

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

$x = 0, x = \pm \sqrt{3}$

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

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

$x = \pm \frac{\sqrt{22}}{2}, x = \pm \sqrt{7}$

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

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

$x equals 25$

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

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

$x equals 4$

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

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

$x = \frac{1}{4}$

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

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

$x = \frac{1}{25}, x = \frac{9}{4}$

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

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

$x = −1, x = −512$

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

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

$x = 8, x = −216$

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

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

$x = \frac{27}{8}, x = - \frac{64}{27}$

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

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

$x = \frac{27}{512}, x = 125$

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

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

$x = 1, x = 49$

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

$6 x^{−2} + 13 x^{−1} + 5 = 0$

$x = −2, x = - \frac{3}{5}$

$15 x^{−2} - 26 x^{−1} + 8 = 0$

$8 x^{−2} - 2 x^{−1} - 3 = 0$

$x = −2, x = \frac{4}{3}$

$15 x^{−2} - 4 x^{−1} - 4 = 0$

Writing Exercises

Explain how to recognize an equation in quadratic form.

Answers will vary.

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?


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