MX Algebra Solve Equations in Quadratic Form

Section 9.4Solve Equations in Quadratic Form

Definition

Before you get started, take this readiness quiz.

1

Factor by substitution: \({y}^{4}-{y}^{2}-20.\)
If you missed this problem, review Example 13.

\(\left({y}^{2}+4\right)\left({y}^{2}-5\right)\)

Definition
2

Factor by substitution: \({(y-4)}^{2}+8(y-4)+15.\)
If you missed this problem, review Example 14.

\((y-1)(y+1)\)

Definition
3

Simplify: ⓐ \({x}^{\frac{1}{2}}·{x}^{\frac{1}{4}}\) ⓑ \({({x}^{\frac{1}{3}})}^{2}\) ⓒ \({({x}^{-1})}^{2}.\)
If you missed this problem, review Example 8.

ⓐ \({x}^{\frac{3}{4}}\) ; ⓑ \({x}^{\frac{2}{3}}\) ; ⓒ \({x}^{-2}\)

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.

Table 1
The image displays the mathematical expression x^4 - 4x^2 - 5, a polynomial equation often seen in algebra and calculus.
The mathematical expression (x^2)^2 - 4(x^2) - 5 is shown, representing a quadratic form in terms of x^2. The 'x' variable and its exponents are highlighted in red.
Let \(u={x}^{2}\) and substitute.The image shows the mathematical expression u^2 - 4u - 5, where 'u' is highlighted in red, indicating it as a variable.
Factor the trinomial.The mathematical expression (u + 1)(u - 5) is displayed on a white background.
Replace u with \({x}^{2}\) .A mathematical expression showing the product of two binomials: (x^2 + 1)(x^2 - 5). The variable 'x' is in red, while the numbers, operators, and parentheses are in black.

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 1How to Solve Equations in Quadratic Form

Solve: \(6{x}^{4}-7{x}^{2}+2=0\)

Notice \(({x}^{2})^{2}={x}^{4},\) so let \(u={x}^{2}\) to rewrite the equation as a quadratic in \(u.\)

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 #1

Solve: \({x}^{4}-6{x}^{2}+8=0\) .

\(x=\sqrt{2},x=\text{-}\sqrt{2},x=2,x=-2\)

Did you get it?
Try It #2

Solve: \({x}^{4}-11{x}^{2}+28=0\) .

\(x=\sqrt{7},x=\text{-}\sqrt{7},x=2,x=-2\)

Did you get it?

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

Solve equations in quadratic form.
  • Identify a substitution that will put the equation in quadratic form.
  • Rewrite the equation with the substitution to put it in quadratic form.
  • Solve the quadratic equation for u.
  • Substitute the original variable back into the results, using the substitution.
  • Solve for the original variable.
  • 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 2

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

Let \(u=x-2\) so the equation becomes a quadratic in \(u.\)

Table 2
A quadratic equation (x-2)^2 + 7(x-2) + 12 = 0 is displayed on a white background, representing a math problem typically solved using substitution.
Prepare for the substitution.Equation showing the expression (x minus 2) squared plus 7 times (x minus 2) plus 12 equals zero.
Let \(u=x-2\) and substitute.A quadratic equation displayed on a white background: u^2 + 7u + 12 = 0. The variable 'u' appears in red, while the numbers, exponents, and operators are in black.
Solve by factoring.

A mathematical equation is displayed against a white background. The equation reads as follows: (u + 3)(u + 4) = 0. Two mathematical equations are displayed: u + 3 = 0, u + 4 = 0, shown in a white background with black text. The image displays two mathematical expressions: u = -3, and u = -4, written in a clear, sans-serif font against a plain white background.
Replace \(u\) with \(x-2.\)Two mathematical equations are displayed: x - 2 = -3 and x - 2 = -4. These are linear equations where a variable 'x' is involved in a subtraction operation resulting in a negative integer.
Solve for \(x.\)Two mathematical equations are displayed on a white background: 'X = -1' followed by a comma, and then 'X = -2'.
Check:

Verification of solutions for a quadratic equation (x-2)^2 + 7(x-2) + 12 = 0, showing that both x=-1 and x=-2 satisfy the equation.
Try It #3

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

\(x=3,x=1\)

Did you get it?
Try It #4

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

\(y=-1,y=1\)

Did you get it?

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: \(x-3\sqrt{x}+2=0.\)

Since \({(\sqrt{x})}^{2}=x,\) let \(u=\sqrt{x}\) to rewrite the equation as a quadratic in \(u.\)

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.

Table 3
A mathematical equation is displayed: x - 3 times the square root of x + 2 = 0. This is a radical equation that can be solved by substitution, letting y = square root of x.
Rewrite the trinomial to prepare for the substitution.A quadratic equation, (sqrt(x))^2 - 3sqrt(x) + 2 = 0, is shown, with the square root of x highlighted in red.
Let \(u=\sqrt{x}\) and substitute.A white background displays the quadratic equation u^2 - 3u + 2 = 0 in horizontal alignment. The variable 'u' is rendered in red, while the numbers, operators, and equals sign are black.
Solve by factoring.

A mathematical equation is displayed on a white background: (u-2)(u-1)=0. Two mathematical equations are displayed: u - 2 = 0, u - 1 = 0, separated by a comma. The equations are presented in a clear, sans-serif font against a plain white background. The image displays mathematical notation 'u = 2, u = 1' in black text against a white background.
Replace u with \(\sqrt{x}.\)Two mathematical equations are displayed: square root of x equals 2, and square root of x equals 1.
Solve for x, by squaring both sides.The text 'x = 4, x = 1' is shown against a white background.
Check:

Checking the solutions x=4 and x=1 for the equation x - 3sqrt(x) + 2 = 0, demonstrating that both values correctly satisfy the equality.
Try It #5

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

\(x=9,x=16\)

Did you get it?
Try It #6

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

\(x=4,x=16\)

Did you get it?

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

Since \({({x}^{\frac{1}{3}})}^{2}={x}^{\frac{2}{3}},\) let \(u={x}^{\frac{1}{3}}\) to rewrite the equation as a quadratic in \(u.\)

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.

Table 4
A mathematical equation is displayed, reading 'x to the power of 2/3 minus 2x to the power of 1/3 minus 24 equals 0'.
Rewrite the trinomial to prepare for the substitution.A mathematical equation is displayed: (x raised to the power of 1/3) squared minus 2 times (x raised to the power of 1/3) minus 24 equals 0.
Let \(u={x}^{\frac{1}{3}}\) and substitute.A quadratic equation is displayed, showing u^2 - 2u - 24 = 0. The variable 'u' and its coefficient '2' in the second term are highlighted in red.
Solve by factoring.

A mathematical equation in factored form is shown against a white background: (u - 6)(u + 4) = 0. Two mathematical equations are displayed: u - 6 = 0 and u + 4 = 0. These linear equations show simple algebraic expressions involving the variable 'u' and constants, both set to zero. Two mathematical expressions are shown: u = 6, and u = -4.
Replace u with \({x}^{\frac{1}{3}}.\)Two mathematical equations are shown: x^(1/3) = 6, and x^(1/3) = -4. These represent two separate conditions for x raised to the power of one-third.
Solve for \(x\) by cubing both sides.
Two algebraic equations are displayed: (x^(1/3))^3 = (6)^3 and (x^(1/3))^3 = (-4)^3, illustrating how to solve for x when involving cube roots and cubing. Two mathematical expressions are displayed on a white background: 'x = 216,' and 'x = -64'.
Check:

This image demonstrates verifying the solutions x=216 and x=-64 for the equation x^(2/3) - 2x^(1/3) - 24 = 0 by substituting them into the equation, both resulting in 0=0.
Try It #7

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

\(x=-8,x=343\)

Did you get it?
Try It #8

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

\(x=81,x=625\)

Did you get it?

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: \(3{x}^{-2}-7{x}^{-1}+2=0.\)

Since \({({x}^{-1})}^{2}={x}^{-2},\) let \(u={x}^{-1}\) to rewrite the equation as a quadratic in \(u.\)

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.

Table 5
A mathematical equation displays '3x^(-2) - 7x^(-1) + 2 = 0' in a simple, clear font against a white background.
Rewrite the trinomial to prepare for the substitution.A mathematical equation is displayed: 3(x⁻¹)² - 7(x⁻¹) + 2 = 0. The variable 'x' is shown in red, with an exponent of -1. The equation features a quadratic form with respect to x⁻¹.
Let \(u={x}^{-1}\) and substitute.A quadratic equation is displayed on a white background: 3u^2 - 7u + 2 = 0. The coefficients for the u-squared and u terms, '3u' and '7u', are highlighted in red, contrasting with the black numbers and operators.
Solve by factoring.A mathematical equation where the product of two binomials, (3u - 1) and (u - 2), is set equal to zero.
Two separate linear equations, 3u - 1 = 0 and u - 2 = 0, are shown in white text against a dark background, representing a mathematical problem to solve for the variable 'u'.
Two mathematical expressions are displayed: u = 1/3, and u = 2.
Replace u with \({x}^{-1}.\)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 \({x}^{-1}=\frac{1}{x}.\)Two mathematical expressions are displayed: 'x = 3,' and 'x = 1/2'. These represent potential solutions or values for the variable x.
Check:

Two algebraic problems demonstrate the verification of solutions for the equation 3x^-2 - 7x^-1 + 2 = 0, with x=3 and x=1/2 both resulting in 0=0.
Try It #9

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

\(x=\frac{4}{3}x=2\)

Did you get it?
Try It #10

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

\(x=\frac{2}{5},x=\frac{3}{4}\)

Did you get it?
Media

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

Key Concepts

Section Exercises

Practice Makes Perfect

Solve Equations in Quadratic Form

In the following exercises, solve.

4

\({x}^{4}-7{x}^{2}+12=0\)

\(x=\pm \sqrt{3},x=\pm 2\)

5

\({x}^{4}-9{x}^{2}+18=0\)

6

\({x}^{4}-13{x}^{2}-30=0\)

\(x=\pm \sqrt{15},x=\pm \sqrt{2}i\)

7

\({x}^{4}+5{x}^{2}-36=0\)

8

\(2{x}^{4}-5{x}^{2}+3=0\)

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

9

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

10

\(2{x}^{4}-7{x}^{2}+3=0\)

\(x=\pm \sqrt{3},x=\pm \frac{\sqrt{2}}{2}\)

11

\(3{x}^{4}-14{x}^{2}+8=0\)

12

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

\(x=-1,x=12\)

13

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

14

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

\(x=-\frac{5}{3},x=0\)

15

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

16

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

\(x=0,x=\pm \sqrt{3}\)

17

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

18

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

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

19

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

20

\(x-\sqrt{x}-20=0\)

\(x=25\)

21

\(x-8\sqrt{x}+15=0\)

22

\(x+6\sqrt{x}-16=0\)

\(x=4\)

23

\(x+4\sqrt{x}-21=0\)

24

\(6x+\sqrt{x}-2=0\)

\(x=\frac{1}{4}\)

25

\(6x+\sqrt{x}-1=0\)

26

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

\(x=\frac{1}{25},x=\frac{9}{4}\)

27

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

28

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

\(x=-1,x=-512\)

29

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

30

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

\(x=8,x=-216\)

31

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

32

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

\(x=\frac{27}{8},x=-\frac{64}{27}\)

33

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

34

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

\(x=\frac{27}{512},x=125\)

35

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

36

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

\(x=1,x=49\)

37

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

38

\(6{x}^{-2}+13{x}^{-1}+5=0\)

\(x=-2,x=-\frac{3}{5}\)

39

\(15{x}^{-2}-26{x}^{-1}+8=0\)

40

\(8{x}^{-2}-2{x}^{-1}-3=0\)

\(x=-2,x=\frac{4}{3}\)

41

\(15{x}^{-2}-4{x}^{-1}-4=0\)

Writing Exercises

42

Explain how to recognize an equation in quadratic form.

Answers will vary.

43

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?