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Study Guides > College Algebra CoRequisite Course

Solving Other Types of Equations

Learning Outcomes

  • Solve polynomial equations.
  • Solve absolute value equations.
We have used factoring to solve quadratic equations, but it is a technique that we can use with many types of polynomial equations which are equations that contain a string of terms including numerical coefficients and variables. When we are faced with an equation containing polynomials of degree higher than 2, we can often solve them by factoring.

recall the zero-product property

Solving by factoring depends on the zero-product property which states that if a⋅b=0a \cdot b=0, then a=0a=0 or b=0b=0, where aa and bb are real numbers or algebraic expressions. In other words if the product of two expressions equals zero, then at least one of the expressions must equal zero. We may apply the zero-product property to polynomial equations in the same way we did to quadratic equations.

A General Note: Polynomial Equations

A polynomial of degree n is an expression of the type
anxn+an−1xn−1+⋅⋅⋅+a2x2+a1x+a0{a}_{n}{x}^{n}+{a}_{n - 1}{x}^{n - 1}+\cdot \cdot \cdot +{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}
where n is a positive integer and an,…,a0{a}_{n},\dots ,{a}_{0} are real numbers and an≠0{a}_{n}\ne 0. Setting the polynomial equal to zero gives a polynomial equation. The total number of solutions (real and complex) to a polynomial equation is equal to the highest exponent n.

Try It

[ohm_question]34186[/ohm_question]

Solving an Absolute Value Equation

An absolute value equation is an equation in which the variable of interest is contained within absolute value bars. Absolute value is defined as a distance. That is, the bars are used to designate that the number inside the absolute value represents its distance from zero on the number line. For example, to solve an equation such as ∣2x−6∣=8,\mid2x - 6\mid=8, notice that the absolute value will be equal to 8 if the quantity inside the absolute value bars is 88 or −8-8. This leads to two different equations we can solve independently.
2x−6=8 or 2x−6=−82x=142x=−2x=7x=−1\begin{array}{lll}2x - 6=8\hfill & \text{ or }\hfill & 2x - 6=-8\hfill \\ 2x=14\hfill & \hfill & 2x=-2\hfill \\ x=7\hfill & \hfill & x=-1\hfill \end{array}
Checking both solutions by evaluating them in the original equation shows that both 77 and −1-1 both make the statement true. In applications, we may need to identify numbers or points on a line that are a specified distance from a given reference point. We can use an absolute value to accomplish such a task.

A General Note: Absolute Value Equations

The absolute value of x is written as ∣x∣|x|. It has the following properties:
If x≥0, then ∣x∣=x.If x<0, then ∣x∣=−x.\begin{array}{l}\text{If } x\ge 0,\text{ then }|x|=x.\hfill \\ \text{If }x<0,\text{ then }|x|=-x.\hfill \end{array}
For real numbers AA and BB, an equation of the form ∣A∣=B|A|=B, with B≥0B\ge 0, will have solutions when A=BA=B or A=−BA=-B. If B<0B<0, the equation ∣A∣=B|A|=B has no solution. An absolute value equation in the form ∣ax+b∣=c|ax+b|=c has the following properties:
If c<0,∣ax+b∣=c has no solution.If c=0,∣ax+b∣=c has one solution.If c>0,∣ax+b∣=c has two solutions.\begin{array}{l}\text{If }c<0,|ax+b|=c\text{ has no solution}.\hfill \\ \text{If }c=0,|ax+b|=c\text{ has one solution}.\hfill \\ \text{If }c>0,|ax+b|=c\text{ has two solutions}.\hfill \end{array}

How To: Given an absolute value equation, solve it

  1. Isolate the absolute value expression on one side of the equal sign.
  2. If c>0c>0, write and solve two equations: ax+b=cax+b=c and ax+b=−cax+b=-c.

Example: Solving Absolute Value Equations

Solve the following absolute value equations:
  1. ∣6x+4∣=8|6x+4|=8
  2. ∣3x+4∣=−9|3x+4|=-9
  3. ∣3x−5∣−4=6|3x - 5|-4=6
  4. ∣−5x+10∣=0|-5x+10|=0

Answer: a. ∣6x+4∣=8|6x+4|=8 Write two equations and solve each:

6x+4=8 or 6x+4=−86x=46x=−12x=23x=−2\begin{array}{lllllllll}6x+4=8\hfill & \text{ or } & 6x+4=-8\hfill & \\ 6x=4\hfill & \hfill & 6x=-12\hfill & \\ x=\frac{2}{3}\hfill& \hfill & x=-2\hfill \end{array}

The two solutions are x=23x=\frac{2}{3}, x=−2x=-2. b. ∣3x+4∣=−9|3x+4|=-9 There is no solution as an absolute value cannot be negative. c. ∣3x−5∣−4=6|3x - 5|-4=6 Isolate the absolute value expression and then write two equations.
∣3x−5∣−4=6∣3x−5∣=103x−5=103x−5=−103x=153x=−5x=5x=−53\begin{array}{lll}\hfill & |3x - 5|-4=6\hfill & \hfill \\ \hfill & |3x - 5|=10\hfill & \hfill \\ \hfill & \hfill & \hfill \\ 3x - 5=10\hfill & \hfill & 3x - 5=-10\hfill \\ 3x=15\hfill & \hfill & 3x=-5\hfill \\ x=5\hfill & \hfill & x=-\frac{5}{3}\hfill \end{array}
There are two solutions: x=5x=5, x=−53x=-\frac{5}{3}. d. ∣−5x+10∣=0|-5x+10|=0 The equation is set equal to zero, so we have to write only one equation.
−5x+10=0−5x=−10x=2\begin{array}{ccc}-5x+10=0\hfill & \\ -5x=-10\hfill & \\ x=2\hfill \end{array}
There is one solution: x=2x=2.

recall properties of equality

When solving absolute value equations, the absolute value expression must be isolated on one side of the equation before setting up the two cases to remove the absolute value bars. Use the properties of equality to isolate the absolute value expression, but avoid multiplying into or dividing from any expression inside the bars. Ex. use the properties of equality and the techniques of solving linear equations to isolate the absolute value expression in the equation 5−2∣3x−4∣=−75-2\mid3x-4\mid=-7, then solve the equation.

Answer: 5−2∣3x−4∣=−7−2∣3x−4∣=−12subtract the 5 to the right-hand side∣3x−4∣=6 divide by -2\begin{align}5-2\mid3x-4\mid &= -7 \\ -2\mid3x-4\mid &= -12 \quad \text{subtract the 5 to the right-hand side} \\ \mid3x-4\mid &= 6 \qquad \ \text{divide by -2} \end{align}   Solution  x=−23,x=103 x=- \dfrac{2}{3}, x= \dfrac{10}{3}

Try It

Solve the absolute value equation: ∣1−4x∣+8=13|1 - 4x|+8=13.

Answer: x=−1x=-1, x=32x=\frac{3}{2}

[ohm_question]60839[/ohm_question]

Other Types of Equations

There are many other types of equations in addition to the ones we have discussed so far. We will see more of them throughout the text. Here, we will discuss equations that are in quadratic form and rational equations that result in a quadratic.

Solving Equations in Quadratic Form

Equations in quadratic form are equations with three terms. The first term has a power other than 2. The middle term has an exponent that is one-half the exponent of the leading term. The third term is a constant. We can solve equations in this form as if they were quadratic. A few examples of these equations include x4−5x2+4=0,x6+7x3−8=0{x}^{4}-5{x}^{2}+4=0,{x}^{6}+7{x}^{3}-8=0, and x23+4x13+2=0{x}^{\frac{2}{3}}+4{x}^{\frac{1}{3}}+2=0. In each one, doubling the exponent of the middle term equals the exponent on the leading term. We can solve these equations by substituting a variable for the middle term.

A General Note: Quadratic Form

If the exponent on the middle term is one-half of the exponent on the leading term, we have an equation in quadratic form which we can solve as if it were a quadratic. We substitute a variable for the middle term to solve equations in quadratic form.

How To: Given an equation quadratic in form, solve it

  1. Identify the exponent on the leading term and determine whether it is double the exponent on the middle term.
  2. If it is, substitute a variable, such as u, for the variable portion of the middle term.
  3. Rewrite the equation so that it takes on the standard form of a quadratic.
  4. Solve using one of the usual methods for solving a quadratic.
  5. Replace the substitution variable with the original term.
  6. Solve the remaining equation.

Example: Solving a Fourth-Degree Equation in Quadratic Form

Solve this fourth-degree equation: 3x4−2x2−1=03{x}^{4}-2{x}^{2}-1=0.

Answer: This equation fits the main criteria: that the power on the leading term is double the power on the middle term. Next, we will make a substitution for the variable term in the middle. Let u=x2u={x}^{2}. Rewrite the equation in u.

3u2−2u−1=03{u}^{2}-2u - 1=0
Now solve the quadratic.
3u2−2u−1=0(3u+1)(u−1)=0\begin{array}{ll}3{u}^{2}-2u - 1=0\hfill & \\ \left(3u+1\right)\left(u - 1\right)=0\hfill \end{array}
Solve for u.
3u+1=03u=−1u=−13x2=−13x=±i13\begin{array}{lllll}3u+1=0\hfill & \\ 3u=-1\hfill & \\ u=-\frac{1}{3}\hfill & \\ {x}^{2}=-\frac{1}{3}\hfill & \\ x=\pm i\sqrt{\frac{1}{3}}\hfill \end{array}
Replace u with its original term.
u−1=0u=1x2=1x=±1\begin{array}{llll}u - 1=0\hfill & \\ u=1\hfill & \\ {x}^{2}=1\hfill & \\ x=\pm 1\hfill \end{array}
The solutions are x=±i13x=\pm i\sqrt{\frac{1}{3}} and x=±1x=\pm 1.

Try It

Solve using substitution: x4−8x2−9=0{x}^{4}-8{x}^{2}-9=0.

Answer: x=−3,3,−i,ix=-3,3,-i,i

Example: Solving an Equation in Quadratic Form Containing a Binomial

Solve the equation in quadratic form: (x+2)2+11(x+2)−12=0{\left(x+2\right)}^{2}+11\left(x+2\right)-12=0.

Answer: This equation contains a binomial in place of the single variable. The tendency is to expand what is presented. However, recognizing that it fits the criteria for being in quadratic form makes all the difference in the solving process. First, make a substitution letting u=x+2u=x+2. Then rewrite the equation in u.

u2+11u−12=0(u+12)(u−1)=0\begin{array}{ll}{u}^{2}+11u - 12=0\hfill & \\ \left(u+12\right)\left(u - 1\right)=0\hfill \end{array}
Solve using the zero-factor property and then replace u with the original expression.
u+12=0u=−12x+2=−12x=−14\begin{array}{llll}u+12=0\hfill & \\ u=-12\hfill & \\ x+2=-12\hfill & \\ x=-14\hfill \end{array}
The second factor results in
u−1=0u=1x+2=1x=−1\begin{array}{llll}u - 1=0\hfill & \\ u=1\hfill & \\ x+2=1\hfill & \\ x=-1\hfill \end{array}
We have two solutions: x=−14x=-14, x=−1x=-1.

Try It

Solve: (x−5)2−4(x−5)−21=0{\left(x - 5\right)}^{2}-4\left(x - 5\right)-21=0.

Answer: x=2,x=12x=2,x=12

[ohm_question]1883[/ohm_question]

Solving Rational Equations Resulting in a Quadratic

Earlier, we solved rational equations. Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution.

Recall division by zero

Dividing zero by a number results in zero.

0a=0\dfrac{0}{a} = 0.

But we cannot divide by zero. A zero in a denominator results in an undefined expression.

b0= undefined\dfrac{b}{0}=\text{ undefined}

When solving rational equations in which a variable is contained in a denominator, always declare the value of the variable which would result in an undefined equation so that you won't forget to exclude it from the solution set.

Example: Solving a Rational Equation Leading to a Quadratic

Solve the following rational equation: −4xx−1+4x+1=−8x2−1\frac{-4x}{x - 1}+\frac{4}{x+1}=\frac{-8}{{x}^{2}-1}.

Answer: We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, x2−1=(x+1)(x−1){x}^{2}-1=\left(x+1\right)\left(x - 1\right). Then, the LCD is (x+1)(x−1)\left(x+1\right)\left(x - 1\right). Next, we multiply the whole equation by the LCD.

(x+1)(x−1)[−4xx−1+4x+1]=[−8(x+1)(x−1)](x+1)(x−1)−4x(x+1)+4(x−1)=−8−4x2−4x+4x−4=−8−4x2+4=0−4(x2−1)=0−4(x+1)(x−1)=0x=−1x=1\begin{array}{cccccccc}\left(x+1\right)\left(x - 1\right)\left[\frac{-4x}{x - 1}+\frac{4}{x+1}\right]=\left[\frac{-8}{\left(x+1\right)\left(x - 1\right)}\right]\left(x+1\right)\left(x - 1\right) \\ -4x\left(x+1\right)+4\left(x - 1\right)=-8 \\ -4{x}^{2}-4x+4x - 4=-8 \\ -4{x}^{2}+4=0 \\ -4\left({x}^{2}-1\right)=0 \\ -4\left(x+1\right)\left(x - 1\right)=0 \\ x=-1 \\ x=1 \end{array}
In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution.

Try It

Solve 3x+2x−2+1x=−2x2−2x\frac{3x+2}{x - 2}+\frac{1}{x}=\frac{-2}{{x}^{2}-2x}.

Answer: x=−1x=-1, x=0x=0 is not a solution.

[ohm_question]3496[/ohm_question]

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