```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
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 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.
x 
2
36
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
• 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.
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x2 + 7x =18
x = -9
(-9)2 + 7(-9) =18
81 – 63 = 18
 18 = 18
x2 + 7x =18
x=2
(2)2 + 7(2) =18
4 + 14 = 18
18 = 18 (the answers check)
Try these two “different” problems:
2x2 + 4x =6
16x2 = 8x
• 2x2 + 4x – 6 = 0
x2
2( + 2x – 3)= 0
2(x+3) (x-1) = 0
Divide by 2
(x + 3) (x – 1) = 0
x + 3 = 0; x-1 =0
x = -3; x = 1
2x2
or
+ 4x – 6 = 0
or (2x + 6) (x – 1) = 0
or (2x + 6) (x – 1) = 0
or 2x+6 = 0 ; x-1 = 0
or 2x = -6 ; x = 1
x = -3 ; x = 1
The solutions are -3 and 1
Both methods yield the same answers
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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
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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 = -5/2
The solutions are 5/2 and -5/2
Both methods yield the same answers
equation.
3x2 – 24=0
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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
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