1. Algebra 1
Section 12.5

2. Quadratic Equations
Any quadratic equation can be solved by completing the square.
Example 1 derives a formula we call the quadratic formula.

3. Theorem If ax2 + bx + c = 0, where a ≠ 0, then -b ± b2 – 4ac x = 2a
This equation is called the quadratic formula.

4. The Quadratic Formula -b ± b2 – 4ac x = 2a
Be sure to write your equation in the standard form before determining the values of a, b, and c.

5. Example 2 Use the quadratic formula to find the solution set for
x2 + 5x – 6 = 0.
a = 1
b = 5
c = -6

6. Example 2 a = 1 -b ± b2 – 4ac b = 5 x = c = -6 2a -5 ± 52 – 4(1)(-6)
-5 ± 52 – 4(1)(-6)
2(1)
x =
-5 ± 25 – (-24) 2

7. Example 2 x = -5 ± 25 – (-24) 2 x = -5 ± 49 2 = -5 ± 7 2 -5 + 7 2
-5 ± 25 – (-24) 2
x =
-5 ± 49 2 =
-5 ± 7 2
-5 + 7 2
-5 – 7 2
= 1
= -6
{-6, 1}

8. Solving Quadratic Equations
Write the equation in standard form: ax2 + bx + c = 0.
Identify the values of a, b, and c.
Substitute these values into the quadratic formula and simplify.

9. Example 3 Use the quadratic formula to solve 3y2 – 4y = 8.
b = -4
c = -8

10. Example 3 a = 3 -b ± b2 – 4ac b = -4 x = c = -8 2a
-(-4) ± (-4)2 – 4(3)(-8)
2(3)
x =
4 ± 16 – (-96) 6

11. Example 3 x = 4 ± 16 – (-96) 6 x = 4 ± 112 6 = 4 ± 4 7 6 = 2(2 ± 2 7)
4 ± 16 – (-96) 6
x =
4 ± 112 6 =
4 ± 4 7 6 =
2(2 ± 2 7)
2(3) =
2 ± 2 7 3

12. Example 4 Let x = the length of each square to be cut out
24 – 2x = length of bottom of box
18 – 2x = width of bottom of box
265 = (24 – 2x)(18 – 2x)
265 = 432 – 84x + 4x2

13. Example 4
265 = 432 – 84x + 4x2
0 = 4x2 – 84x + 167
a = 4, b = -84, c = 167
x =
-(-84) ± (-84)2 – 4(4)(167)
2(4)
x =
-b ± b2 – 4ac 2a

14. Example 4 -(-84) ± (-84)2 – 4(4)(167) x = 2(4) 84 ± 7056 – 2672 x = 8
-(-84) ± (-84)2 – 4(4)(167)
2(4)
x =
84 ± – 2672 8
x =
84 ± 4384 8

15. Example 4
x =
84 ± 4384 8
x = 8
≈ 18.8 in.
x =
84 – 8
≈ 2.2 in.

16. Homework: pp