Transcript The Quadratic Formula.

The Quadratic Formula.
 b  b  4ac
x
2a
2
What Does
The Formula Do ?
The Quadratic formula allows you to find the roots of a quadratic
equation (if they exist) even if the quadratic equation does not
factorise.
The formula states that for a quadratic equation of the form :
ax2 + bx + c = 0
The roots of the quadratic equation are given by :
 b  b  4ac
x
2a
2
Example 1
Use the quadratic formula to solve the equation :
x 2 + 5x + 6= 0
Solution:
x 2 + 5x + 6= 0
a=1 b=5 c=6
 b  b 2  4ac
x
2a
 5  52  (4  1  6)
x
21
 5  25  (24)
x
2
5 1
x
2
x
51
or
2
x
 5 1
2
x = - 2 or x = - 3
These are the roots of the equation.
Example 2
Use the quadratic formula to solve the equation :
8x 2 + 2x - 3= 0
Solution:
8x 2 + 2x - 3= 0
a = 8 b = 2 c = -3
 b  b  4ac
x
2a
2
 2  22  (4  8  3)
x
2 8
 2  4  (96)
x
16
 2  100
x
16
 2  10
x
16
or
 2  10
x
16
x = ½ or x = - ¾
These are the roots of the equation.
Example 3
Use the quadratic formula to solve the equation :
8x 2 - 22x + 15= 0
22  (484 (480)
x
16
Solution:
2
8x - 22x + 15= 0
a = 8 b = -22 c = 15
 b  b 2  4ac
x
2a
 (22)  (22)2  (4  8  15)
x
2 8
22 4
x
16
x
22  2
16
or
x
22  2
16
x = 3/2 or x = 5/4
These are the roots of the equation.
Example 4
Use the quadratic formula to solve for x
2x 2 +3x - 7= 0
Solution:
2x 2 + 3x – 7 = 0
a=2 b=3 c=-7
 3  9  ( 56)
x
4
 3  65
x
4
 b  b 2  4ac
x
2a
 3  32  (4  2  7)
x
2 2
These are the roots of the equation.
• The Quadratic Formula
–The solutions of a quadratic equation of
the form ax2 + bx + c = 0, where a ≠ 0,
are given by the following formula:
b  b  4ac
x
2a
2
Two Rational Roots
x 2  12 x  28
x 2  12 x  28  0
• Solve
by using the Quadratic Formula.
b  b  4ac
x
2a
2
( 12)  ( 12) 2  4(1)( 28)
x
2(1)
12  16
12  16
or x 
x
2
2
12  144  112
x
2
 14
 2
12  256
x
2
Solutions
are
-2
and
14.
12  16
x
2
• Solve
One Rational Root
x  22 x  121  0
2
by using the Quadratic Formula.
b  b  4ac
x
2a
2
22  (22)  4(1)(121)
x
2(1)
2
22  0
x
2
22
x
or 11
2
• The solution is 11 .
• Solve
Irrational Roots
2x  4x  5  0
2
by using the Quadratic Formula.
b  b  4ac
x
2a
2
4  (4)  4(2)( 5)
x
2(2)
2
4  56
x
4
or 2  14
2
2  14
• The exact solutions are
2
and 2  14
2
.
• Solve
Complex Roots
x  4 x  13
2
by using the Quadratic Formula.
b  b  4ac
x
2a
2
( 4)  ( 4)  4(1)(13)
x
2(1)
2
4  36
x
2
4  6i
x
2
x  2  3i
• The exact solutions are
2  3i
and
2  3i
.
Roots and the Discriminant
• The Discriminant = b2 – 4ac
• The value of the Discriminant can be used to
determine the number and type of roots of a
quadratic equation.
Discriminant
Value of Discriminant
Type and Number of
Roots
b2 – 4ac > 0;
b2 – 4ac is a perfect
square.
2 real, rational roots
b2 – 4ac > 0;
2 real, irrational roots
b2 – 4ac is not a perfect
square.
b2 – 4ac = 0
1 real, rational root
b2 – 4ac < 0
2 complex roots
Example of Graph of
Related Function
Describe Roots
• Find the value of the discriminant for each quadratic equation.
Then describe the number and type of roots for the equation.
2
• A. 9 x  12 x  4  0
a  9, b  12, c  4
b2  4ac  (12)2  4(9)(4)
 144  144  0
The discriminant is 0, so there is one rational root.
2
• B. 2 x  16 x  33  0
a  2, b  16, c  33
b2  4ac  (16)2  4(2)(33)
 256  264
 8
The discriminant is negative, so there are two complex roots.
Describe Roots
• Find the value of the discriminant for each quadratic equation.
Then describe the number and type of roots for the equation.
• C. 5x 2  8 x  1  0
a  5, b  8, c  1
b2  4ac  (8)2  4(5)(1)  64  20  44
The discriminant is 44, which is not a perfect square. Therefore
there are two irrational roots.
D.
7 x  15x  4  0
2
a  7, b  15, c  4
2
2
b  4ac  (7)  4(15)(4)
 17

289

49

240
The discriminant is 289, which is a perfect square.
there are two rational roots.
2
Therefore,