Algebra 1 Section 12.5

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

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

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.

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

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}

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.

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

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

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

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

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

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 ± 7056 – 2672 8 x = 84 ± 4384 8

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

Homework: pp. 505-506