1. Algebra 1
Section 12.1

2. Solving Quadratic Equations
Linear equations are first- degree equations, since the variable is only to the first power.
Quadratic equations are second-degree equations.

3. Definition A quadratic equation is an equation of the second degree.
Its standard form is
ax2 + bx + c = 0, where a, b, and c are real numbers and
a ≠ 0.

4. Zero Product Property
In the equation xy = 0, we know that either x = 0 or y = 0, or possibly both equal zero.
We use this property to solve quadratic equations.
If pq = 0, then p = 0 or q = 0.

5. Example 1 Solve (x – 5)(x + 2) = 0. x – 5 = 0 or x + 2 = 0
There are two solutions:
x = 5, -2.

6. Solving Quadratic Equations
Many quadratic equations must be factored before you can apply the Zero Product Property.
Be sure to have a zero on one side of the equation before factoring.

7. Example 2 Solve x2 + 8x + 12 = 0. (x + 2)(x + 6) = 0
x + 2 = 0 or x + 6 = 0
x = x = -6
Be sure to check your solutions in the original equation.

8. Example 3 Solve 8x2 – 16x = 0. 8x(x – 2) = 0 8x = 0 or x – 2 = 0

9. Solving Quadratic Equations
Write the equation in standard form: ax2 + bx + c = 0.
Factor the polynomial side of the equation.
Set each factor that contains a variable equal to zero using the Zero Product Property.

10. Solving Quadratic Equations
Solve the resulting equations.
Check your solutions in the original equation.

11. Example 4 Solve 2x2 + 8x – 42 = 0. 2(x2 + 4x – 21) = 0
2 ≠ 0 or x + 7 = 0 or x – 3 = 0
x = x = 3

12. Example 5 Solve 3x2 = 147. 3x2 – 147 = 0 3(x2 – 49) = 0
3 ≠ 0 or x – 7 = 0 or x + 7 = 0
x = x = -7

13. Solving Quadratic Equations
Quadratic equations will have at most two real number solutions.
The Zero Product Property can also be applied to equations with higher degrees after they have been factored.

14. Example 6 Solve 4x3 = 14x2 + 8x. 4x3 – 14x2 – 8x = 0
2x = 0 or 2x + 1 = 0 or x – 4 = 0
x = x = -½ x = 4

15. Solving Quadratic Equations
Dividing both sides of an equation by a variable expression causes a solution to be lost.
Some word problems require you to write and solve a quadratic equation.

16. Example 7 Let w = the width w + 5 = the length
Since (length)(width) = area,
w(w + 5) = 204
w2 + 5w = 204
w2 + 5w – 204 = 0

17. Example 7 w2 + 5w – 204 = 0 (w + 17)(w – 12) = 0
w + 17 = 0 or w – 12 = 0
w = w = 12
Since the width cannot be negative, w = 12.

18. Example 7 The width [w] is 12 ft. The length [w + 5] is 17 ft.
Check: The area is (12 ft)(17ft) = 204 ft2.

19. Homework: pp