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Guias de estudo > College Algebra

Solving Other Types of Equations

Learning Objectives

  • 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.

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.
 

Solve an Absolute Value Equation

Next, we will learn how to slve an absolute value equation. To solve an equation such as ∣2x−6∣=8|2x - 6|=8, we 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}
Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point.

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=86x+4=−86x=46x=−12x=23x=−2\begin{array}{ll}6x+4\hfill&=8\hfill& 6x+4\hfill&=-8\hfill \\ 6x\hfill&=4\hfill& 6x\hfill&=-12\hfill \\ x\hfill&=\frac{2}{3}\hfill& x\hfill&=-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}{l}-5x+10\hfill&=0\hfill \\ -5x\hfill&=-10\hfill \\ x\hfill&=2\hfill \end{array}
There is one solution: x=2x=2.

Try It

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

Answer: x=−1x=-1, x=32x=\frac{3}{2}

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.

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}{l}\left(x+1\right)\left(x - 1\right)\left[\frac{-4x}{x - 1}+\frac{4}{x+1}\right]\hfill&=\left[\frac{-8}{\left(x+1\right)\left(x - 1\right)}\right]\left(x+1\right)\left(x - 1\right)\hfill \\ -4x\left(x+1\right)+4\left(x - 1\right)\hfill&=-8\hfill \\ -4{x}^{2}-4x+4x - 4\hfill&=-8\hfill \\ -4{x}^{2}+4\hfill&=0\hfill \\ -4\left({x}^{2}-1\right)\hfill&=0\hfill \\ -4\left(x+1\right)\left(x - 1\right)\hfill&=0\hfill \\ x\hfill&=-1\hfill \\ x\hfill&=1\hfill \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.

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