1.4 Quadratic Equations

1.4 Quadratic Equations
Chapter 1 Equations and Inequalities

Concepts and Objectives
⚫ Quadratic Equations
⚫ Solve quadratic equations, finding all solutions
⚫ Cubic Equations
⚫ Solve the sum or difference of two cubes

Quadratic Equations
⚫ A quadratic equation is an equation that can be written
in the form
where a, b, and c are real numbers, with a  0. This is
standard form.
⚫ A quadratic equation can be solved by factoring,
graphing, completing the square, or by using the
quadratic formula.
⚫ Graphing and factoring don’t always work, but
completing the square and the quadratic formula will
always provide the solution(s).
+ + =2
0ax bx c

Factoring Quadratic Equations
⚫ Factoring works because of the zero-factor property:
⚫ If a and b are complex numbers with ab = 0, then
a = 0 or b = 0 or both.
⚫ To solve a quadratic equation by factoring:
⚫ Put the equation into standard form (= 0).
⚫ If the equation has a GCF, factor it out.
⚫ Using the method of your choice, factor the quadratic
expression.
⚫ Set each factor equal to zero and solve both factors.

Example: Solve by factoring.− − =2
2 15 0x x

Example: Solve by factoring.
The solution set is
− − =2
2 15 0x x
= = − = −2, 1, 15a b c –30
–1
–6 5− + − =2
2 6 5 15 0x x x
( ) ( )− + − =2 3 5 3 0x x x
( )( )+ − =2 5 3 0x x
+ = − =2 5 0 or 3 0x x
= −
5
, 3
2
x
5
, 3
2
 
− 
 

Square Root Property
⚫ If x2 = k, then
⚫ Both solutions are real if k > 0 and often written as {±k}
⚫ Both solutions are imaginary if k < 0, and written as
⚫ If k = 0, there is only one distinct solution, 0.
orx k x k= = −
 i k

Square Root Property (cont.)
Example: What is the solution set?
⚫ x2 = 17
⚫ x2 = ‒25
⚫ ( )
2
4 12x − =

Square Root Property (cont.)
Example: What is the solution set?
⚫ x2 = 17
⚫ x2 = ‒25
⚫ ( )
2
4 12x − =
 17
 5i
4 12
4 2 3
x
x
− = 
= 
25 5x i=  − = 
 4 2 3
17x = 
Remember to simplify
any radicals!

Completing the Square
⚫ As the last example shows, we can use the square root
property if x is part of a binomial square.
⚫ It is possible to manipulate the equation to produce a
binomial square on one side and a constant on the other.
We can then use this method to solve the equation. This
method is called completing the square.

Completing the Square (cont.)
Solving a quadratic equation (ax2 + bx + c = 0) by
completing the square:
⚫ If a  1, divide everything on both sides by a.
⚫ Isolate the constant (c) on the right side of the equation.
⚫ Add ½b2 to both sides.
⚫ Factor the now-perfect square on the left side.
⚫ Use the square root property to complete the solution.

Completing the Square (a = 1)
Example: Solve x2 ‒ 4x ‒ 14 = 0 by completing the square.

Completing the Square (a = 1)
Example: Solve x2 ‒ 4x ‒ 14 = 0 by completing the square.
( )
2
2
2 2 2
2
4 14 0
4 14
4 14
1
2
2 2
2 18
2
2 8
3
x x
x x
x x
x
x
x
− − =
− =
+ = +
=
− = 
−

−
=

Completing the Square (a  1)
Example: Solve 4x2 + 6x + 5 = 0 by completing the square.

Completing the Square (a  1)
Example: Solve 4x2 + 6x + 5 = 0 by completing the square.
2
2
2
22 2
2
2
4 6 5 0
3 5
0 Divide by 4
2 4
3 5
2 4
3 3 5 3 1 3
Add to each side
2 4 4 4 2 2
3 11
4 16
3 11
4 16
3 11
4 4
x x
x x
x x
x x
x
x
x i
+ + =
+ + =
+ = −
      
+ + = − +      
      
 
+ = − 
 
+ =  −
= − 
3 11
The solution set is
4 4
i
  
−  
  

Quadratic Formula
⚫ The solutions of the quadratic equation ,
where a  0, are
⚫ Example: Solve
+ + =2
0ax bx c
−  −
=
2
4
2
b b ac
x
a
= −2
2 4x x

Quadratic Formula
⚫ Example: Solve = −2
2 4x x
− + =2
2 4 0x x a cb 4, ,2 1== =−

Quadratic Formula
⚫ Example: Solve
The solution set is
= −2
2 4x x
− + =2
2 4 0x x a cb 4, ,2 1== =−
( ) ( ) ( )( )
( )
x
2
2
2
1 1 4
2
4−  −
=
− −
 −  −
= =
1 1 32 1 31
4 4

=
1 31
4
i   
 
  
1 31
4 4
i

Cubic Equations
⚫ We will mainly be working with cubic equations that are
the sum or difference of two cubes:
a3  b3 = 0
⚫ Equations of this form factor as
⚫ To solve this, set each factor equal to zero and solve.
(Use the Quadratic Formula or Completing the Square
for the quadratic factor.)
( )( ) + =2 2
0a b a ab b

Cubic Equations
Example: Solve 8x3 + 125 = 0

Cubic Equations
Example: Solve 8x3 + 125 = 0 ( ) + =
3 3
2 5 0x
( )( )+ − + =2
2 5 4 10 25 0x x x

Cubic Equations
Example: Solve 8x3 + 125 = 0 ( ) + =
3 3
2 5 0x
( )( )+ − + =2
2 5 4 10 25 0x x x
+ = − + =2
2 5 0 or 4 10 25 0x x x

Cubic Equations
Example: Solve 8x3 + 125 = 0 ( ) + =
3 3
2 5 0x
( )( )+ − + =2
2 5 4 10 25 0x x x
+ = − + =2
2 5 0 or 4 10 25 0x x x
+ =
= −
= −
2 5 0
2 5
5
2
x
x
x

Cubic Equations
Example: Solve 8x3 + 125 = 0 ( ) + =
3 3
2 5 0x
( )( )+ − + =2
2 5 4 10 25 0x x x
+ = − + =2
2 5 0 or 4 10 25 0x x x
+ =
= −
= −
2 5 0
2 5
5
2
x
x
x
( ) ( )( )
( )
 − −
=
2
10 10 4 4 25
2 4
x
 −  
= = =
10 300 10 10 3 5 5 3
8 8 4
i i

Cubic Equations
Example: Solve 8x3 + 125 = 0 ( ) + =
3 3
2 5 0x
( )( )+ − + =2
2 5 4 10 25 0x x x
+ = − + =2
2 5 0 or 4 10 25 0x x x
+ =
= −
= −
2 5 0
2 5
5
2
x
x
x
( ) ( )( )
( )
 − −
=
2
10 10 4 4 25
2 4
x
 −  
= = =
10 300 10 10 3 5 5 3
8 8 4
i i
  
−  
  
5 5 5 3
,
2 4 4
i
The solution set is

Classwork
⚫ College Algebra
⚫ Page 119: 14-30, page 109: 38-54, page 89: 44-54
(skip 48) (all even)

1.4 Quadratic Equations

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1.4 Quadratic Equations