This document discusses solving quadratic equations by factoring. It begins by defining a quadratic equation in standard form and explaining the zero factor property. It then provides examples of solving quadratic equations through factoring and setting each factor equal to zero. Finally, it demonstrates solving word problems by setting up and solving the resulting quadratic equation.
1. 5.3 Solving Quadratic Equations by Factoring
A quadratic equation is written in the Standard
Form,
2
ax + bx + c = 0
where a, b, and c are real numbers and a ≠ 0.
Examples:
x − 7 x + 12 = 0
x ( x + 7) = 0
2
3x + 4 x = 15
2
(standard form)
12. 5.3 Solving Quadratic Equations by Factoring
Zero Factor Property:
If a and b are real numbers and if ab = 0 ,
then a = 0 or b = 0 .
Examples:
x ( x + 7) = 0
x=0
x+7 =0
x=0
x = −7
13. 5.3 Solving Quadratic Equations by Factoring
Zero Factor Property:
If a and b are real numbers and if ab = 0 ,
then a = 0 or b = 0 .
Examples:
( x − 10 ) ( 3x − 6 ) = 0
x − 10 = 0
3x − 6 = 0
x − 10 + 10 = 0 + 10 3 x − 6 + 6 = 0 + 6
3x 6
x = 10
=
x=2
3x = 6
3 3
14. 11.6 – Solving Quadratic Equations by Factoring
Solving Quadratic Equations:
1) Write the equation in standard form.
2) Factor the equation completely.
3) Set each factor equal to 0.
4) Solve each equation.
5) Check the solutions (in original equation).
16. 5.3 Solving Quadratic Equations by Factoring
If the Zero Factor
Property is not used,
then the solutions will
be incorrect
x − 3 x = 18
x ( x − 3) = 18
x = 18
x − 3 = 18
( 18)
2
324 − 54 = 18
270 ≠ 18
2
x − 3 + 3 = 18 + 3
x = 21
− 3 ( 1 8 ) = 18
( 21)
2
− 3 ( 21) = 18
441 − 63 = 18
378 ≠ 18
17. 5.3 Solving Quadratic Equations by Factoring
x ( x − 4) = 5
x − 4x = 5
2
x − 4x − 5 = 0
2
( x + 1) ( x − 5) = 0
x +1 = 0
x −5 = 0
x = −1
x=5
18. 5.3 Solving Quadratic Equations by Factoring
x ( 3x + 7 ) = 6
( x + 3) ( 3 x − 2 ) = 0
3x + 7 x = 6
x + 3 = 0 3x − 2 = 0
x = −3
3x = 2
x=2
3
2
3x + 7 x − 6 = 0
Factors of 3 :
1, 3
Factors of 6 :
1, 6 2, 3
2
19. 5.3 Solving Quadratic Equations by Factoring
9 x − 24 x = −16
2
9 x − 24 x + 16 = 0
2
( 9 and 16 are perfect squares )
( 3x − 4 ) ( 3x − 4 ) = 0
3x − 4 = 0
3x = 4
4
x=
3
20. 5.3 Solving Quadratic Equations by Factoring
2 x − 18 x = 0
2
2x ( x − 9 ) = 0
3
2x ( x + 3) ( x − 3) = 0
2x = 0 x + 3 = 0 x − 3 = 0
x=3
x=0
x = −3
21. 5.3 Solving Quadratic Equations by Factoring
( x + 3) ( 3 x
− 20 x − 7 ) = 0
Factors of 3 : 1, 3 Factors of 7 : 1, 7
2
( x + 3) ( x − 7 ) ( 3x + 1) = 0
x+3= 0
x = −3
x−7 = 0
x=7
3x + 1 = 0
3 x = −1
1
x=−
3
22. 5.3 Quadratic Equations and Problem Solving
A cliff diver is 64 feet above the surface of the
water. The formula for calculating the height (h)
of the diver after t seconds is: h = −16t 2 + 64.
How long does it take for the diver to hit the
surface of the water?
2
0 = −16t + 64
2
0 = −16 ( t − 4 )
0 = −16 ( t + 2 ) ( t − 2 )
t+2=0
t = −2
t−2=0
t = 2 seconds
23. 11.7 – Quadratic Equations and Problem Solving
The square of a number minus twice the number is
63. Find the number.
x is the number.
x −2x = 63
2
x − 2 x − 63 = 0
Factors of 63 : 1, 63 3, 21 7, 9
2
( x + 7 ) ( x − 9)
=0
x+7 =0
x−9 = 0
x = −7
x=9
24. 5.3 Quadratic Equations and Problem Solving
The length of a rectangular garden is 5 feet more than
its width. The area of the garden is 176 square feet.
What are the length and the width of the garden?
l ×w = A The width is w. The length is w+5.
( w + 5) w = 176
( w − 11) ( w + 16 )
w + 5w = 176
w − 11 = 0
w = 11
2
w2 + 5w − 176 = 0
Factors of 176 :
1, 176 2, 88 4, 44
8, 22 11, 16
w = 11 feet
=0
w + 16 = 0
w = −16
l = 11 + 5
l = 16 feet
25. 5.3 Quadratic Equations and Problem Solving
Find two consecutive odd numbers whose product is
23 more than their sum?
x + 2.
Consecutive odd numbers: x
x ( x + 2 ) = ( x + x + 2 ) +23 x + 5 = 0
2
x + 2 x = 2 x + 25
x = −5
x 2 + 2 x − 2 x = 2 x + 25 − 2 x −5 + 2 = −3
2
x = 25
−5, − 3
2
x − 25 = 25 − 25
x 2 − 25 = 0
( x + 5) ( x − 5) = 0
x −5 = 0
x=5
5+ 2 = 7
5, 7
26. 5.3 Quadratic Equations and Problem Solving
The length of one leg of a right triangle is 7 meters less than
the length of the other leg. The length of the hypotenuse is 13
meters. What are the lengths of the legs? ( Pythagorean Th.)
a 2 + b2 = c 2
a = x b = x − 7 c = 13 2 ( x + 5 ) ( x − 12 ) = 0
2
2
x + ( x − 7 ) = 132
x − 12 = 0
x+5 = 0
x 2 + x 2 − 14 x + 49 = 169
x = −5
x = 12
2 x 2 − 14 x − 120 = 0
a = 12 meters
2
2 ( x − 7 x − 60 ) = 0
b = 12 − 7 = 5 meters
Factors of 60 : 1, 60 2, 30
3, 20 4, 15 5, 12 6, 10