1. Algebra 1
Section 12.2

2. Solving Quadratic Equations
Another method of solving equations of the form x2 – c = 0 is to isolate the squared term and then take the square root of each side of the equation.

3. Example 1
Solve x2 – 64 = 0.
x2 = 64
x2 = 64
|x| = 8
x = ±8

4. Square Root Property If n ≥ 0 and x2 = n, then x = ±n.
If n > 0, then x has two real solutions.
If n = 0, then x has one real solution.

5. Square Root Property
If n < 0 and x2 = n, then x has no real solutions.

6. Example 2
Solve 2x2 – 24 = 0.
2x2 = 24
x2 = 12
x = ± 12
x = ± 2 3

7. Example 3 Solve (x + 2)2 = 36. x + 2 = ± 6 x = -2 ± 6 x = 4, -8
Did you check each solution in the original equation?

8. Solving Equations of the Form ax2 – c = 0
Isolate the squared expression.
Apply the Square Root Property.
Simplify the radical and solve each equation.

9. Example 4 Solve (x – 5)2 = 21. x – 5 = ± 21 x = 5 ± 21 x ≈ 5 ± 4.58
The solution set is {0.42, 9.58}.

10. Example 5 Solve 7x2 + 6 = 2x2 + 8. 5x2 + 6 = 8 5x2 = 2 x2 = x = ± 2 5

11. Example 5
x = ± 2 5
x = ± 2 5 5 •
= ± 10 5

12. Solving Quadratic Equations
Solving x2 + 7 = 0 yields the solutions x = ± -7.
The square root of -7 is not a real number.

13. Solving Quadratic Equations
For now, solve the problem, leaving the negative number under the radical; and state that there are no real solutions.

14. The Free Fall Equation
Assuming negligible air resistance, the height of a dropped object can be modeled by the function h(t) = -16t2 + hi, where t is the time in seconds and hi is the initial height in feet.

15. Example 6 h(t) = -16t2 + hi 0 = -16t2 + 150 16t2 = 150 t2 = t = ± 150

16. Only t ≈ 3.06 sec is a reasonable answer.
Example 6
t = ±
150 16
t = ±
5 6 4
≈ ± 3.06
Only t ≈ 3.06 sec is a reasonable answer.

17. Homework: pp