2.2
Objectives:
1. Solve equations by:
A. Factoring
B. Square Root of Both Sides
C. Completing the Square
D. Quadratic Formula
2. Solve equations in quadratic form.
Definition of a Quadratic Equation
An equation that can be written in the form
ax2 + bx + c = 0 with real constants a, b, & c, with a ≠ 0.
Techniques to solve a quadratic equation:
 Techniques that sometimes work
Factoring
Taking the square root of both sides of an equation
 Techniques that always work
Completing the Square
Using the Quadratic Formula
Example #1a
Solving by Factoring
Solve by Factoring:
x  5x  6
2
x 2  5x  6  0
x  6x  1  0
x  6  0 x 1 0
x  6 x  1
Example #1b
Solving by Factoring
Solve by Bottoms Up Method:
4x  22x  12
2
Don’t forget to first factor out the GCF, if necessary.
4 x 2  22 x  12  0
2 x 2  11x  6
2 2 x 2  11x  6  0
x 2  11x  12
x  12 x  1


2 x  6 2 x  1  0
x  6  0 2x  1  0
x  6 2 x  1
1
x
2
12 
1

 x   x  
2 
2

x  62 x  1
To factor this trinomial
you must first multiply
a by c, factor normally,
divide both factors by a
& reduce, and finally
bring “bottoms up”.
Example #2
Solving ax2 = b
Solve by Taking the Square Root of Both Sides:
4x  8
2
4x2 8

4
4
x2  2
x 2
x  1.414
Example #3
Solving a(x − h)2 = k
Solve by Taking the Square Root of Both Sides:
5(x  3) 2  15
5 x  32 15

5
5
x  32  3
x  32
 3
x 3 3
x 3 3
x  4.732 or
x  1.268
Example #4
Solving a Quadratic Equation by Completing the Square
Solve by Completing the Square:
8x  24x  3  0
2
8 x 2  24 x  3 0

8
8
3
2
x  3x   0
8
Completing the square only works
when the coefficient of x2 is a 1.
Always divide every term by the
leading coefficient before attempting
to complete the square.
Example #4
Solving a Quadratic Equation by Completing the Square
Solve by Completing the Square:
Remember after
completing the
square, add it to
both sides of the
equation.
2
 b    3
  

2
2
  

9

4
2
2
3
3
15
x  3x   0

x   
8
2
8

3
2
x  3x  
3
15
8
x 
2
8
9
3 9
2
x  3x    
4
8 4
3
15
x 
2
2
8
3  15

x  
x  2.869 or x  0.131
2
8

2
The Quadratic Formula
x
b 
b  4ac
2a
2
Example #5
Solving a Quadratic Equation by Using the Quadratic Formula
Solve by Using the Quadratic Formula:
x  13  10x
2
Quadratic equations
must be in Standard
Form ax2 + bx + c = 0
before using the
Quadratic Formula.
x 2  13  10 x
x 2  10 x  13  0
x
10 
 10 2  41 13
21
10  100  52
2
10  152

2
 10  2 38
2
 5  38
 11.2 or

 1.2
The Discriminant
b  4ac
2
The discriminant is used to determine the number of
solutions without solving the problem.
Discriminant is Positive
Two Real Solutions
Discriminant is Zero
One Real Solution
Discriminant is Negative
No Real Solution
Example #6
Determining the Number of Solutions by Using the Discriminant
Find the Number of Solutions:
4x  x  2
2
4x  x  2  0
2
1
 44 2 
 1  32
 31
2
−31 is NOT a solution
to the equation.
Since the discriminant is
negative, this means that
there is no real solution
to the equation.
Example #7
Polynomials in Quadratic Form
Solving an Equation in Quadratic Form:
3x  5x  2  0
4
2
Compare this polynomial equation with the quadratic form:
ax2 + bx + c = 0
This polynomial can be factored using Bottoms Up!
x 4  5x 2  6
x
2

 2 x2  3
3x

 2 2  2 3 
 x   x  
3 
3

3x
2


 2 x2  1
2


 2 x2  1  0
3x 2  2  0
x2  1  0
3x 2  2
x2  1
2
x
3
x  1