1. Copyright 2013, 2010, 2007, 2005, Pearson, Education, Inc.

2. Solving Quadratic Equations by Factoring
6.6
Solving Quadratic Equations by Factoring

3. Quadratic Equation Quadratic Equation
A quadratic equation is one that can be written in the form ax2 + bx + c = 0, where a, b and c are real numbers and a ≠ 0.
The form ax2 + bx + c = 0 is referred to as standard form.

4. Zero-Factor Property Zero-Factor Property
If a and b are real numbers and if ab = 0, then a = 0 or b = 0.

5. Example Solve: (x – 4)(x + 2) = 0 x – 4 = 0 or x + 2 = 0
You can check the solutions on your own.

6. Solving Quadratic Equations
To Solve Quadratic Equations by Factoring
Step 1: Write the equation in standard form.
Step 2: Factor the quadratic completely.
Step 3: Set each factor containing a variable equal to 0.
Step 4: Solve the resulting equations.
Step 5: Check each solution in the original equation.

7. Example
Solve: x2 – 5x = 24
First write the quadratic equation in standard form.
x2 – 5x – 24 = 0
Now factor.
x2 – 5x – 24 = (x – 8)(x + 3) = 0
Set each factor equal to 0.
x – 8 = 0 or x + 3 = 0, which will simplify to
x = 8 or x = –3
Continued

8. Example (cont) Check both possible answers in the original equation.
82 – 5(8) = 64 – 40 = True
(–3)2 – 5(–3) = 9 – (–15) = True
So the solutions are 8 or –3. ?

9. Example Solve: 4x(8x + 9) = 5 4x(8x + 9) = 5 32x2 + 36x = 5
8x – 1 = 0 or 4x + 5 = 0
8x = x = – 5
x =
x =
The solutions are
and
Continued

10. Example (cont) Check both possible answers in the original equation.
True ?
True ?
The solutions are
and

11. Example
Solve:
Replace x with each solution in the original equation. The solutions all check.

12. Finding x-intercepts
Recall that in Chapter 3, we found the x-intercept of linear equations by letting y = 0 and solving for x.
The same method works for x-intercepts in quadratic equations.
Note: When the quadratic equation is written in standard form, the graph is a parabola opening up (when a > 0) or down (when a < 0), where a is the coefficient of the x2 term.
The intercepts will be where the parabola crosses the x-axis.

13. Example Find the x-intercepts of the graph of y = 4x2 + 11x + 6.
The equation is already written in standard form, so we let y = 0, then factor the quadratic in x.
0 = 4x2 + 11x + 6 = (4x + 3)(x + 2)
We set each factor equal to 0 and solve for x.
4x + 3 = 0 or x + 2 = 0
4x = –3 or x = –2
x = –¾ or x = –2
So the x-intercepts are the points (–¾, 0) and (–2, 0).