This document provides an overview of solving quadratic equations by factoring when possible and using the quadratic formula when factoring is not possible. It defines the standard form of a quadratic equation as y = ax2 + bx + c and explains that the roots are the x-intercepts where y = 0. Examples are given of factoring a quadratic equation to find the roots and of using the quadratic formula. The key steps of the quadratic formula are outlined and further examples are worked through to demonstrate solving various quadratic equations.

1.  Intro: We already know the
standard form of a quadratic
equation is:
y = ax2 + bx + c
 The coefficients are: a , b, c
 The variables are: y, x

2.  The ROOTS (or
solutions) of a
polynomial are
its x-intercepts
 The x-intercepts
occur where y =
0.
Roots

3.  Example: Find the
roots: y = x2 + x - 6
 Solution: Factoring:
y = (x + 3)(x - 2)
 0 = (x + 3)(x - 2)
 The roots are:
 x = -3; x = 2
Roots

4.  After centuries of
work,
mathematicians
realized that as long
as you know the
coefficients, you can
find the roots of the
quadratic. Even if it
doesn’t factor!
y  ax2
 bx  c, a  0
x 
b  b2
 4ac
2a

5. Solve: y = 5x2
 8x  3
x 
b  b2
 4ac
2a
a  5, b  8, c  3
x 
(8)  (8)2
 4(5)(3)
2(5)
x 
8  64  60
10
x 
8  4
10
x 
8  2
10

6. x 
8  2
10
x 
8  2
10

10
10
 1
x 
8  2
10

6
10

3
5
Roots

7. y  5(1)2
 8(1)  3
y  5  8  3
y  0
y  5 3
5
 
2
 8 3
5
  3
y  5 9
25
  24
5
 3
y  45
25
  24
5
  3
y  9
5
  24
5
  15
5
 
y  0
Plug in your
answers for x.
If you’re right,
you’ll get y = 0.

8. Solve : y  2x2
 7x  4
a  2, b  7, c  4
x 
b  b2
 4ac
2a
x 
(7)  (7)2
 4(2)(4)
2(2)
x 
7  49  32
4
x 
7  81
4
x 
7  9
4
x 
2
4

1
2
x 
16
4
 4

9. Remember: All the terms must be on one
side BEFORE you use the quadratic
formula.
•Example: Solve 3m2 - 8 = 10m
•Solution: 3m2 - 10m - 8 = 0
•a = 3, b = -10, c = -8

10.  Solve: 3x2 = 7 - 2x
 Solution: 3x2 + 2x - 7 =
0
 a = 3, b = 2, c = -7
x 
b  b2
 4ac
2a
x 
(2)  (2)2
 4(3)(7)
2(3)
x 
2  4  84
6
x 
2  88
6
x 
2  4• 22
6
x 
2  2 22
6
x 
1 22
3

11.  Watch this:
http://www.youtube.com/watch?v=jGJrH49Z2ZA