4.1: Quadratic Equations and Solving by Factoring

Learning Objectives
Note

Before you get started, take this readiness quiz.

  1. Solve: \(5y−3=0\).
    If you missed this problem, review Example 2.3.1.
  2. Solve: \(10a=0\).
    If you missed this problem, review Example 2.2.1.
  3. Combine like terms: \(12 x^-6 x+4 x\).
    If you missed this problem, review Example 1.3.37.
  4. Factor \(n^-9 n^-22 n\) completely.
    If you missed this problem, review Example 7.3.10.

We have already solved linear equations, equations of the form \(a x+b y=c\). In linear equations, the variables have no exponents. Quadratic equations are equations in which the variable is squared. Listed below are some examples of quadratic equations:

\[x^+5 x+6=0 \quad 3 y^+4 y=10 \quad 64 u^-81=0 \quad n(n+1)=42\]

The last equation doesn’t appear to have the variable squared, but when we simplify the expression on the left we will get \(n^+n\).

The general form of a quadratic equation is \(a x^+b x+c=0\),with \(a \neq 0\).

QUADRATIC EQUATION

An equation of the form \(a x^+b x+c=0\) is called a quadratic equation.

To solve quadratic equations we need methods different than the ones we used in solving linear equations. We will look at one method here and then several others in a later chapter.

Solve Quadratic Equations Using the Zero Product Property

We will first solve some quadratic equations by using the Zero Product Property. The Zero Product Property says that if the product of two quantities is zero, it must be that at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.

ZERO PRODUCT PROPERTY

We will now use the Zero Product Property, to solve a quadratic equation.

Example \(\PageIndex\): How to Use the Zero Product Property to Solve a Quadratic Equation

Solution

This table gives the steps for solving (x + 1)(x – 4) = 0. The first step is to set each factor equal to 0. Since it is a product equal to 0, at least one factor must equal 0. x + 1 = 0 or x – 4 = 0.The next step is to solve each linear equation. This gives two solutions, x = −1 or x = 4.The last step is to check both answers by substituting the values for x into the original equation.

Try It \(\PageIndex\)

Answer