# Zero – product property

## Presentation on theme: "Zero – product property"— Presentation transcript:

Zero – product property
5-5 Quadratic equations Zero – product property

Zero product property After factoring a trinomial you can set each binomial equal to zero and solve for your variable Example: Solve x2 + 7x =18 First, put all terms on the left side of the equation and set the trinomial equal to zero x2 + 7x – 18 = 0 Factor the trinomial into 2 binomials (x + 9) (x-2) = 0 Set each binomial equal to zero and solve each part. x + 9 = 0 x – 2 = 0 x = x = 2 The solutions are -9 and 2.

Solving by factoring square roots
Example: Solve 5x = 0 Rewrite the equation so the squared term is on the left side and the constant is on the right 5x2 = 180 Isolate the variable Divide by 5 to isolate the x-squared term x2 = 180/5 x2 = 36 Take the square root of each side to solve for x x2 = 36 x =+ 6

Check the solutions back in the original problem. Sometimes one of the solutions won’t check. If the solution doesn’t check, it is thrown out. x2 + 7x =18 x2 + 7x =18 x = -9 x = 2 (-9)2 + 7(-9) =18 (2)2 + 7(2) =18 81 – 63 = = 18 18 = = 18 (the answers check)

Try these two “different” problems:
2x2 + 4x =6 16x2 = 8x 2x2 + 4x – 6 = 0 2( x2 + 2x – 3)= 0 or 2x2 + 4x – 6 = 0 2(x+3) (x-1) = or (2x + 6) (x – 1) = 0 Divide by 2 (x + 3) (x – 1) = 0 or (2x + 6) (x – 1) = 0 x + 3 = 0; x-1 =0 or 2x+6 = 0 ; x-1 = 0 x = -3; x = 1 or 2x = -6 ; x = 1 x = -3 ; x = 1 The solutions are -3 and 1 Both methods yield the same answers Check your answers in the original equation. 16x2 - 8x = 0 8x (2x – 1 ) = 0 8x = x – 1 = 0 x = x = 1 x = ½ The solutions are 0 and ½

Try these two “different” problems:
4x =0 3x2 – 24=0 4x2 = 25 Divide by 4 x2 = 25/4 x = + 5/2 Or 4x2 – 25 = 0 (rewrite the problem as the diff. of 2 squares) (2x-5) (2x+5) = 0 2x-5 = 0 and 2x+5 = 0 x=5/ x = -5/2 The solutions are 5/2 and -5/2 Both methods yield the same answers Check your answers in the original equation. 4x2 = 25 Divide by 4 x2 = 25/4 x = + 5/2 Or 4x2 – 25 = 0 (rewrite the problem as the diff. of 2 squares) (2x-5) (2x+5) = 0 2x-5 = 0 and 2x+5 = 0 x=5/ x = -5/2 The solutions are 5/2 and -5/2 Both methods yield the same answers Check your answers in the original equation. 3x2 = 24 Divide by 3 x2 = 8 x = + 8 The solutions are 8 and - 8

Homework Chapter 5 packet ; 5-5 w/s 1, 5, 8, 9, 18, 28, 30