The Quadratic Formula and
the Discriminant
Algebra 2HN
Derive the quadratic formula from ax2 +
bx + c = 0 a≠ 0
ax  bx c  0
2
General form of a quadratic equation.
2
ax bx c 0
  
a
a a a
c
2 bx
x   0
a a
Divide all by a
Simplify
2
1
c
b
x  x    
2
a
a
2
Subtract c/a on both sides.
Multiply by ½ and square the result.
Derive the quadratic formula from ax2 +
bx + c = 0 a≠ 0
2
b
c  b 
 b 
x  x     
a
a  2a 
 2a 
2
2
2
b 
b
c

x   2 
2a 
4a
a

Add the result to both sides.
2
Simplify
2
2
b
b
c 4a


x   2  
2a 
4a
a 4a

2
b
b
 4ac


x  
2
2a 
4a

Multiply by common denominator
2
Simplify
Derive the quadratic formula from ax2 +
bx + c = 0 a≠ 0
b 
b  4ac

x  
2
2
a
4
a


2
2
b
b  4ac
x

2a
2a
Square root both sides
2
Simplify
b
b
b
b  4ac
x



2a 2a
2a
2a
2
 b  b  4ac
x
2a
2
Simplify
Common denominator/subtract
from both sides
Quadratic Formula
The solutions of a quadratic equation of
the form ax2 + bx + c with a ≠ 0 are given
by this formula:
 b  b  4ac
x
2a
2
MEMORIZE!!!!
Ex. 1: Solve t2 – 3t – 28 = 0
a = 1 b = -3 c = -28
 b  b 2  4ac
x
2a
 (3)  (3) 2  4(1)( 28)
x
2(1)
3  9  112
x
2
3  121
x
2
3  11
x
2
3  11 14
x

7
2
2
3  11  8
x

 4
2
2
There are 2 distinct roots—Real and rational.
Ex. 1: Solve t2 – 3t – 28 = 0
CHECK:
t2 – 3t – 28 = 0
72 – 3(7) – 28 = 0
49 – 21 – 28 = 0
49 – 49 = 0 
CHECK:
t2 – 3t – 28 = 0
(-4)2 – 3(-4) – 28 = 0
16 + 12 – 28 = 0
28 – 28 = 0 
Ex. 1: Solve t2 – 3t – 28 = 0 -- GRAPH
fx  = x 2 -3x-28
4
2
-5
5
-2
-4
-6
-8
-10
10
15
Ex. 2: Solve x2 – 8x + 16 = 0
a = 1 b = -8 c = 16
 b  b 2  4ac
x
2a
 (8)  (8) 2  4(1)(16)
x
2(1)
8  64  64
x
2
8 0
x
2
80
x
2
8
x 4
2
There is 1 distinct root—Real and rational.
Ex. 2: Solve x2 – 8x + 16 = 0
CHECK:
x2 – 8x + 16 = 0
(4)2 – 8(4) + 16 = 0
16 – 32 + 16 = 0
32 – 32 = 0 
There is 1 distinct root—Real and rational.
Ex. 2: Solve Solve x2 – 8x + 16 = 0 -GRAPH
8
6
4
2
5
-2
-4
-6
10
15
20
Ex. 3: Solve 3p2 – 5p + 9 = 0
a = 3 b = -5 c = 9
 b  b 2  4ac
x
2a
 (5)  (5) 2  4(3)(9)
x
2(3)
5  25  108
x
6
There is 2 imaginary roots.
5   83
x
6
5  i 83
x
6
5  i 83
x
6
5  i 83
x
6
Ex. 3: Solve 3p2 – 5p + 9 = 0
20
18
16
14
12
NOTICE THAT THE PARABOLA DOES NOT
TOUCH THE X-AXIS.
10
8
6
4
2
5
-2
10
15
20
25
30
Note:
These three examples demonstrate a
pattern that is useful in determining the
nature of the root of a quadratic equation.
In the quadratic formula, the expression
under the radical sign, b2 – 4ac is called
the discriminant. The discriminant tells the
nature of the roots of a quadratic
equation.
DISCRIMINANT
b  4ac
2
 The discriminant will tell you about the nature of the
roots of a quadratic equation.
Equation
t2 – 3t – 28 = 0
Value of the
discriminant
Roots
b2 – 4ac =
(-3)2 – 4(1)(-28) = 121
{7, - 4}
x2 – 8x + 16 = 0 b2 – 4ac =
{0}
Nature of
roots
2 real
roots
1 real root
(-8)2 – 4(1)(16) = 0
3p2 – 5p + 9 = 0 b2 – 4ac =
(-5)2 – 4(3)(9) = -83
x
5  i 83
6
2 complex
roots
Ex. 4: Find the value of the discriminant of
each equation and then describe the nature of
its roots.
2x2 + x – 3 = 0
a = 2 b = 1 c = -3
b2 – 4ac = (1)2 – 4(2)(-3)
= 1 + 24
= 25
The value of the discriminant is positive and
a perfect square, so 2x2 + x – 3 = 0 has
two real roots and they are rational.
Ex. 5: Find the value of the discriminant of
each equation and then describe the nature of
its roots.
x2 + 8 = 0
a=1 b=0 c=8
b2 – 4ac = (0)2 – 4(1)(8)
= 0 – 32
= – 32
The value of the discriminant is negative, so
x2 + 8 = 0 has two imaginary/complex
roots.