Solving Quadratic Equations
Using the Quadratic Formula
MA.912.A.7.2 Solve quadratic
equations over the real numbers by
factoring and by using the quadratic
formula.
The Quadratic Formula
The solutions of a quadratic equation written in
Standard Form, ax2 + bx + c = 0, can be found by
Using the Quadratic Formula.
b  b  4ac
x
2a
2
Click on the link below to view a song to help you
memorize it.
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Deriving the Quadratic Formula by
Completing the Square.
ax  bx  c  0

a
a
b
c
2
x  x 0
a
a
b
c
2
x  x
a
a
2
Divide both
sides by “a”.
Subtract constant from
both sides.
Deriving the Quadratic Formula by Completing
the Square.
 b 
b
c  b  Complete
      The Square
x  x  
2a 
a
a 2a 
2
2
Factor the

b 
c
b
Perfect Square
x      2
Trinomial
  2a 
a 4a
2
2
2

b  b  4ac
x   
2
 2a 
4a
2
2
Simplify expression
on the left side by
finding the LCD
Deriving the Quadratic Formula by Completing
the Square.

b 
x   
 2a 
b  4ac
2
4a
2
b
x

2a
2
b  4ac
2
4a
2
b
b  4ac

x

2a
2a
2
Take the square
root of both sides
Don't Forget :
x2  x
Solve absolute value/
Simplify radical
Deriving the Quadratic Formula by Completing
the Square.
b
b  4ac
x

2a
2a
Isolate x
b
b2  4ac
x   b  b  4ac
2a
x  2a
Simplify
2
2
22a
b  b  4ac
x
2a
Congratulations!
You have derived
The Quadratic Formula
#1 Solve using the quadratic formula.
3x  7x  2  0
2
a  3, b  7, c  2
75
x
6
 b  b 2  4ac
x
2a
 (7)  (7) 2  4(3)( 2)
x
2(3)
7  49  24
x
6
7  25
x
6
12
x
6
2
x
6
1
x  2 or x =
3
Graph y  3x  7x  2
2

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
#1 Solve by factoring
3x  7x  2  0
2
3x  1x  2  0
3x  1  0 or x  2  0
1
x = or x  2
3
#2 Solve by factoring
2
2x  4 x  5
2x  4x  5  0
2
2x
2x
5x
1  0
1x
5  0
This quadratic is Prime (will not factor),
The Quadratic Formula must be used!
#2 Solve using the quadratic formula.
2
4  56
2x  4 x  5
x
2x  4x  5  0
2
a  2, b  4, c  5
 b  b  4ac
x
2a
2
Exact
Solution
 (4)  (4)  4(2)( 5)
x
2(2)
2
4  16  40
x
4

Approx
Solution
4
4  2 14
x
4
2  14
x
2
2  3.74
x
2
x  2.87 or x   0.87
Graph y  2x  4 x  5
2

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#3 Solve using the quadratic formula
2
x   5x  7
x  5x  7  0
2
a  1, b  5, c  7
 b  b  4ac
x
2a
 5  25  28
x
2
5  3
x
2
2
 5  5  4(1)(7)
x
2(1)
2
 The 3 is not a real
number, therefore this
equation has
‘NO Real Solution’
Graph
y  x  5x  7
2

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#4 Solve using the quadratic formula
x  64  16 x
2
x  16 x  64  0
2
a  1, b  16, c  64
 b  b 2  4ac
x
2a
x
16 
 16
 4(1)(64)
2(1)
2
16  256  256
x
2
16  0
x
2
x 8
Would factoring work
to solve this equation?
x  8x  8  0
Graph y  x 16  64
2

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#5 Solve using the quadratic formula.
2
2  8
2x 1  x
x
x  2x 1  0
2
a  1, b  2, c  1
 b  b  4ac
x
2a
2
 Exact
Solution
(2)  (2)  4(1)(1) 
x
2(1)
2
2  4  4
x
2

Approx
Solution
2
2  2 2
x
2
x 1  2
x  1  1.41
x  2.41 or x   0.41
#5 What if we move everything to the right
side?
2
2 8
2x 1  x
x
x  2x 1  0
2
a  1, b  2, c  1
 b  b  4ac
x
2a
Exact
2
Solution
(2)  (2)  4(1)(1)
x
2(1)
2
2 44
x
2

Approx
Solution
2
22 2
x
2
x 1  2
x  1  1.41
x  2.41 or x   0.41
y  x  2x 1 & y  x  2x 1
Graph
2
2
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The Discriminant
The expression inside the radical in the
quadratic formula is called the
Discriminant.
b  b  4ac
x
2a
2
The discriminant can be used to
determine the number of solutions that
a quadratic has.
Understanding the discriminant
Discriminant
b 2  4ac
b 2  4ac  0
# of real
solutions
Perfect
square
2 real rational solutions
Not
Perfect
2 real irrational solutions
b  4ac  0
2
b  4ac  0
2
1 real rational
solution
No real
solution
#6 Find the discriminant and describe the
solutions to the equations.
4 y  3 y 1  0
2
a  4, b  3, c  1
discrimina nt  b  4ac
2
2 Real
Rational
Solutions
 (3)  4(4)(1)
2
 9 16
 25
#7 Find the discriminant and describe the
solutions to the equations.
4x  5  x
2
4x  x  5  0
2
a  4, b  1, c  5
discrimina nt  b  4ac
2
 (1)  4(4)(5)

No Real
2
Solutions
 1  80
  79

#8 Find the discriminant and describe the
solutions to the equations.
2x  4x  5
2
2x  4x  5  0
2
a  2, b  4, c  5
discrimina nt  b  4ac
2
 (4)  4(2)(5)
2
2 Real
Rational
Solutions

 16  40
 56
#9 Find the discriminant and describe the
solutions to the equations.
4 x  8x  4
2
4 x  8x  4  0
2
a  4, b  8, c  4
discrimina nt  b  4ac
2


1 Real
Rational
Solution

= (8)  4(4)(4)
2
 64  64
 0