Download Quadratic Formula - Count with Kellogg

5.8 Quadratic Formula
For quadratic equations written in standard form,
the roots can be found using the following formula:
 b  b  4ac
x
2a
2
This is called the Quadratic Formula. It is one of the more
important formulas that you will learn this year.
Memorize it!!!!
Find the roots:
y  x  9 x  14
2
a=1
 b  b 2  4ac
x
2a
b=9
 9  9  4114 c = 14
x
21
2
 9  81  56
x
2
x  2,7
 9  25
x
2
95
95 95
x

or
2
2
2
4
14
  or 
2
2
Find the roots:
y  5x  8x  1
2
  8   8  451
x
25
2
8  64  20
x
10
 b  b 2  4ac
x
2a
a=5
b = -8
c=1
x  0.137, 1.463
8  44
x
10
8  2 11
x
10
8  2 11 8  2 11

or
10
10
Find the roots:
a = -3
b = -5
y  3x  5x  5
c=5
2
  5   5  4 35
x
2 3
2
5  25  60
x
6
5  85
x
6
5  85 5  85

or
6
6
 b  b 2  4ac
x
2a
x  0.703,  2.370
Find the roots:
y  3x  6 x  3
2
  6   6  433
x
23
2
6  36  36
x
6
6 0
x
6
60
60

or
6
6
a=3
b = -6
c=3
x 1
Find the roots:
y  4x  2x  1
 2  2 2  441
x
24
2
 2  4  16
x
8
 2   12
x
8
 2  2 3i
 2  2 3i

or
8
8
a=4
b=2
c=1
 1  1 3i
 1  1 3i
x
or
4
4
Find the roots:
y  7 x  5 x  10
2
5   5  47 10
x
27 
2
5  25  280
x
14
5   255
x
14
a=7
b = -5
c = 10
5  255i
x
14
Discriminants
b2 – 4ac is called the discriminant.
It will allow you to quickly determine how many roots a
particular quadratic equation has.
 b  b 2  4ac
x
2a
Discriminants
In the previous examples, there were three cases.
Case #1: 2 Real solutions: The normal situation for the quadratic
formula is that it gives two real numbers as the solutions.
When the discriminant is positive, there will be 2 real solutions.
There are 2 real solutions
because of what is inside the
square root is positive.
 b  b  4ac
x
2a
2
When you add and subtract a
positive real number, you get
2 different real answers.
Discriminants
Case #2: 1 Real solution: Sometimes both the solutions are the same. It
is then said that there is only one distinct solution.
When the discriminant is zero, there will be 1 real solution.
There is 1 real solution
because of what is inside the
square root is zero.
 b  b  4ac
x
2a
2
When you add and subtract a
zero, you get the same
answer both times.
Discriminants
Case #3: 2 Complex solutions: Sometimes both the solutions are
complex. It is then said that there are no real solutions.
When the discriminant is negative, there will be 2 complex solutions.
There are 2 complex solutions
because of what is inside the
square root is negative. (This
will become an i.)
 b  b  4ac
x
2a
2
When you add and subtract an
imaginary number, you get 2
different complex numbers.
Use the discriminant to determine how many
solutions there will be for each equation.
x2 + 12x + 3 = 0
4x2 - 12x + 9 = 0
b 2  4ac
 144  4(1)(3)
b 2  4ac
 144  4(4)(9)
= 132
=0
Positive Number
Zero
2 Real Solutions
1 Real Solution
Use the discriminant to determine how many
solutions there will be for each equation.
7x2 - x + 2 = 0
24x2 - 14x - 5 = 0
b 2  4ac
 1  4(7)( 2)
b 2  4ac
 196  4(24)( 5)
= -55
= 676
Negative Number
Positive Number
0 Real Solutions
2 Real Solutions
(2 Complex
Solutions)
Use the discriminant to determine how many
solutions there will be for each equation.
16x2 + 40x + 25 = 0
b 2  4ac
 1600  4(16)( 25)
3x2 - 21 = 0
b 2  4ac
 0  4(3)( 21)
=0
= 252
Zero
Positive Number
1 Real Solution
2 Real Solutions
Use the discriminant to determine how many
solutions there will be for each equation.
1 2 1
x  x7 0
2
3
b 2  4ac
1
1
  4( )(7)
9
2
1
125
 14  
9
9
Negative
0 Real Solutions
(2 Complex Solutions)
2
x9  0
5
Not a
quadratic!!!
Don’t use
discriminant!!!