5-5 Quadratic equations

Zero – product property

Solving quadratic equations• 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 = -9 x = 2▫The solutions are -9 and 2.

Solving by factoring square roots

• Example: Solve 5x2 - 180 = 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

x2 36x2 36x2 36

Solving quadratic equations•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 4 + 14 = 18

18 = 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 = 02(x+3) (x-1) = 0 or (2x + 6) (x – 1) = 0Divide 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 = 0x = -3; x = 1 or 2x = -6 ; x = 1

x = -3 ; x = 1

The solutions are -3 and 1

Both methods yield the same answersCheck your answers in the original

equation.

• 16x2 - 8x = 0 • 8x (2x – 1 ) = 0• 8x = 0 2x – 1 = 0• x = 0 2x = 1• x = ½

• The solutions are 0 and ½

Try these two “different” problems:

4x2 - 25 =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=5/2 x = -5/2The solutions are 5/2 and -5/2Both methods yield the same

answersCheck your answers in the

original equation.

• 3x2 = 24 • Divide by 3 • x2 = 8• x = + 8

• The solutions are 8 and - 8

• 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=5/2 x = -5/2The solutions are 5/2 and -5/2Both methods yield the same

answersCheck your answers in the

original equation.

Homework

Chapter 5 packet ; 5-5 w/s1, 5, 8, 9, 18, 28, 30