Download CPM Section 9.4A Quadratic Formula

CPM
Section 9.4A
Quadratic Formula
Thus far we have considered two methods for
solving quadratic function- factoring and using
the square root property. Each of these
methods have their drawbacks however. We
will now consider the……
Quadratic Formula: For the equation , any
solutions are given by the formula
2

b

b
 4ac
X=
2a
153x  13x  40  0
2
a  153
b  13
c  40
13  (13)2  4(153)(40)
2(153)
144
and
306
x
13  24649
306
13  157
306
170
306
8
5
and
17
9
3x  10x  1  0
2
a  3 b  10 c  1
10  (10)2  4(3)(1)
2(3)
10  2 22
6
10  100  12
6
5  1 22
3
10  88
6
10  4  22
6
2 x  3x  5  0
2
a  2 b  3 c  5
3  (3)2  4(2)(5)
2(2)
3  9  40
2(2)
3  31
4
No
Solution.
Can’t have
a negative
number in
the radical
The expression is called the _____________,
discriminant
and it
determines home many solutions (0, 1, or 2) the
quadratic equation has in the set of real numbers.
2
b
If  4ac  0 , the quadratic has ____ solutions in

1 solutions in
If b2  4ac  0 , the quadratic has ____

2 solutions in
If b2  4ac  0, the quadratic has ____

0
Find the number of solutions for the following quadratic equations
a 2  12a  32  0
(12)2  4(1)(32)
144 128
disc  16
2 real sol.
2b2  13b  7
y2  4  4 y
2b 2  13b  7  0
y2  4 y  4  0
(13)2  4(2)(7)
169  56
225
2 real sol.
(4)2  4(1)(4)
16 16
0
1 real sol.