Chapter 10 Section 3
Solving Quadratic Equations
by the Quadratic Formula
Learning Objectives
Solve quadratic equation be the quadratic
formula
Determine the number of solutions to a
quadratic equation using the discriminant
Key Vocabulary
Quadratic formula
Discriminant
Quadratic Formula
b  b  4ac
x
2a
2
 Standard form of a quadratic equation is
ax2 + bx + c = 0
where a is the coefficient of the squared term and b is the
coefficient of the first-degree term and c is the constant.
 It is important to label the a, b, and c with the correct sign
when substituting into the quadratic formula.
Solve the Quadratic Equation by Factoring
x2 + 7x + 10 = 0
(x + 2)(x + 5) = 0
x+2=0
and
x+5=0
x = -2
and
x = -5
Not all equations can be solve by factoring.
Solve the Quadratic Equation by
Completing the Square
x2 + 7x + 10 = 0
( ½ )(7) = 3.5
(3.5)2 = 12.25
x2 + 7x = -10
x2 + 7x + 12.25 = -10 + 12.25
x2 + 7x + 12.25 = 2.25
(x + 3.5)2 = 2.25
Perfect Square Trinomial
( x  3.5)  2.25
2
x + 3.5 = ±1.5
x = -3.5 ±1.5
x = -3.5+1.5 and
x = -2
and
x = -3.5 – 1.5
x = -5
To Solve a Quadratic Equation
by the Quadratic Formula
1. Write the equation in standard form to
determine the values of a, b, and c.
2. Substitute the values for a, b and c from the
equation in standard form into the quadratic
formula and evaluate.
3. Check by placing back into original equation.
Solve using the Quadratic Formula
x2
+ 7x + 10 = 0
(7)  (7) 2  4(1)(10)
x
2(1)
Already in standard form
a=1
b=7
c = 10
b  b2  4ac
x
2a
x
7  49  40
2
x
7  9
2
x
7  3 4

 2
2
2

x
7  3
2
and
x
7  3 10

 5
2
2
x = -2
x = -5
Solve using the Quadratic Formula
18x2
– 3x – 1 = 0
Already in standard form
a = 18
b = -3
c = -1
b  b  4ac
x
2a
2
(3)  (3) 2  4(18)( 1)
x
2(18)
x
3  9  72
36
x
3  81
36
x
3  9 12 1


36
36 3

39
36
x
1
3
and
and
x
1
x
6
3  9 6
1


36
36
6
Solve using the Quadratic Formula
5n2 + 2n - 1 = 0
(2)  (2) 2  4(5)(1)
x
2(5)
Already in standard form
x
a=5
b=2
c = -1
b  b2  4ac
x
2a
2  4  20
10
2  24
x
10

2  24
10
2  4 6 2  2 6
x

10
10
1  6
x
5
Solve using the Quadratic Formula
x2
= 4x – 1
(4)  (4) 2  4(1)(1)
x
2(1)
Put in standard form
x
4  16  4
2
x
4  12
2
x
4 4 3 42 3

2
2
x2 – 4x + 1 = 0
a=1
b = -4
c=1
b  b2  4ac
x
2a
x  2 3
Solve using the Quadratic Formula
4x2 = 2x – 3
Put in standard form
4x2 – 2x + 3 = 0
a=4
b = -2
c=3
(2)  (2) 2  4(4)(3)
x
2(4)
2  4  48
x
8
2  44
x
8
b  b  4ac
x
2a
2
<0
no real number solution
Solve using the Quadratic Formula
m2 = 64
m  64
2
(0)  (0) 2  4(1)(64)
x
2(1)
m2  64
m  8
Put in standard form
x
 256
x
2
m2 +0x – 64 = 0
x
a=1
b=0
c = -64
0  0  256
2
16 16

8
2
2
and
x
16 16

 8
2
2
Solve using the Quadratic Formula
x2 +14x +45 = 0
(14)  (14) 2  4(1)(45)
x
2(1)
Already in standard form
x
14  196  180
2
x
14  16
2
x
14  4
2
x
14  4 10

 5
2
2
a=1
b = 14
c = 45
b  b2  4ac
x
2a
and
x
14  4 18

 9
2
2
x = -5
x = -9
Solve using the Quadratic Formula
12x2 - 4x - 1 = 0
(4)  (4) 2  4(12)(1)
x
2(12)
Already in standard form
x
4  16  48
24
x
4  64
24
x
48
24
x
4  8 12 1


24
24 2
a = 12
b = -4
c = -1
b  b2  4ac
x
2a
and
x
4  8 4 1


24
24 6
1
x
2
and
x
1
6
Solve using the Quadratic Formula
3x2 + 4x - 8 = 0
Already in standard form
a=3
b=4
c = -8
b  b  4ac
x
2a
2
x
(4) 
(4) 2  4(3)( 8)
2(3)
x
4  16  96
6
x
4  112
6
x
4  16 7
6
x
4  4 7
4 4 7
2 2 7




6
6
6
3
3
x
2  2 7
3
Solve using the Quadratic Formula
x2
= 8x – 6
(8)  (8) 2  4(1)(6)
x
2(1)
Put in standard form
x
8  64  24
2
x
8  40
2
x
8  4 10 8  2 10

2
2
x2 – 8x + 6 = 0
a=1
b = -8
c=6
b  b2  4ac
x
2a
x  4  10
Solve using the Quadratic Formula
a2 – 121 = 0
(0)  (0) 2  4(1)( 121)
x
2(1)
Put in standard form
a2 + 0x – 121 = 0
a=1
b=0
c = -121
b  b  4ac
x
2a
x
0  0  484
2
x
 484
2
22
x
2
x
22
 11
2
and
2
x  11
x
 22
 11
2
Solve using the Quadratic Formula
4x2 - 2x + 3 = 0
Already in standard form
a=4
b = -2
c=3
b  b2  4ac
x
2a
(2)  (2) 2  4(4)(3)
x
2(4)
2  4  48
x
8
2  44
x
8
No real solution
<0
Solve an Application Problem using the Quadratic Formula
The length of a rectangle is 1 ft more than three times the width. Find
the dimensions of the rectangle if the area is 30 ft2 , find the length and
width
A = LW
Let x = width
length = 3x + 1
(3x + 1)x = 30
b  b2  4ac
x
2a
a=3
b=1
c = -30
3x2 + x = 30
put in standard form
3x2 + x – 30 = 0
W=3
L = (3)(3) + 1 = 10
(1)  (1) 2  4(3)(30)
x
2(3)
x
1  1  360
6
x
1  361
6
x
1  19
6
x
1  19 18
 3
6
6
Determine the Number of Solutions to a
Quadratic Equation Using the Discriminant
 The expression under the radical is called the discriminant and
can be used to determine the number of solutions
 b2 – 4ac
is called the discriminant
 b2 – 4ac > 0
two distinct real number solutions
0 0
 b2 – 4ac = 0
one real number solution
 b2 – 4ac < 0
no real number solution, negative
No 
4 0 4
 2
2
2
Determine the Number of Solutions to a
Quadratic Equation Using the Discriminant
2x2 + 3x = 5
Put in standard form
2x2 + 3x – 5 = 0
a=2
b=3
c = -5
b2 – 4ac
(3)2 – (4)(2)(-5)
9 – (8)(-5)
9 + 40
49
49 > 0
Two distinct
real number
solutions
Determine the Number of Solutions to a
Quadratic Equation Using the Discriminant
2x2 + x + 3 = 0
Already in standard form
a=2
b=1
c=3
b2 – 4ac
(1)2 – (4)(2)(3)
1 – (8)(3)
1 – 24
-23
-23 < 0
No
real number
solutions
Determine the Number of Solutions to a
Quadratic Equation Using the Discriminant
9x2 - 24 x + 16 = 0
Already in standard form
a=9
b = -24
c = 16
b2 – 4ac
(-24)2 – (4)(9)(16)
256 – (36)(16)
576 – 576
0
0= 0
One
real number
solutions
Determine the Number of Solutions to a
Quadratic Equation Using the Discriminant
2x2 - 3 x - 7 = 0
Already in standard form
a=2
b = -3
c = -7
b2 – 4ac
(-3)2 – (4)(2)(-7)
9 – (8)(-7)
9 + 56
65
65 > 0
Two distinct
real number
solutions
Determine the Number of Solutions to a
Quadratic Equation Using the Discriminant
3x2 = - x + 4
Put in standard form
3x2 + x – 4 = 0
a=3
b=1
c = -4
b2 – 4ac
(1)2 – (4)(3)(-4)
1 – (12)(-4)
1 – 48
-47
-47 < 0
No real
number solutions
Remember
 The equation should be in standard form so that you can
correctly determine a, b, and c.
 Make sure that you label a, b, and c with the correct sign.
 Check your answers by placing them back into the original
equation.
 Use the discriminant to determine the number of solutions
HOMEWORK 10.3
Page 609 – 610:
# 27, 29, 35, 36, 43, 47, 51, 59