Section 1-6: Solving Quadratic Equations
Any equation in the
form
ax2 + bx + c =
0 with
a not equal 0 
is called a quadratic
equation!
A value that
satisfies
this equation is called a root,
zero or solution
of the equation!! We have three methods we can use to solve these
equations:
1) factoring (easy)
2) completing the
square
3) quadratic formula
Factoring review!
Solve each by
factoring
1) x2
+ 4x -5 = 0
Solution: factor
(x
+ 5)(x - 1) = 0
Therefore, x = -5 or x
=
1
2) (3x - 2)((x + 4)
= -11
Solution: Foil
first
3x2 + 10x - 8
= -11
Put in correct
form
3x2 + 10x + 3
= 0
factor
(3x + 1)(x + 3) = 0
Solution:
x = -1/3 or x = -3
Completing the
square!!
Follow the
explanation
and sample problem to review completing the square
1) Use completing
the
square to find the solutions for:
2x2
- 12x - 9 = 0
Solution:
Move the constant to
the
other
side:
2x2 - 12x = 9
Divide by the
coefficient
of x2
x2 - 6x = 9/2
Take half the
coefficient
of x and square: x2
- 6x + 9 = 9/2 + 9
Factor the trinomial
square:
(x - 3)2 = 27/2
Take the square root of
both
sides: 
Simplify the
radical: 
Isolate for x: 
Quadratic Formula
As proved in class
the
quadratice formula is derived by completing the square. Here is
the
formula:
If ax2
+ bx + c = 0 then the roots of the equation are:
Look
familiar?
It better!!
Solve the
following problem
by using the quadratic formula.
2x2
+ 5 = 3x
2x2 - 3x + 5 =
0 (putting
in correct form)
a = 2, b = -3 and c
=
5 Use the formula:
The Discriminant!
The quantity under
the
radical in the quadratic formula can tell us alot about the nature of
the
solutions. Therefore, it is given a special name. The discriminant
is b2
- 4ac
If the discriminant
is
less than zero, then you will be taking the square root of a negative
number
yielding complex solutions. If the discriminant equals zero, you
have one real solution (namely -b/2a) (Where
have I heard that before?). If the
discriminant
is greater than zero, then we have two different real solutions.
This is summarized in the following chart:
| discriminant |
Types of solutions |
| b2 - 4ac < 0 |
Two complex conjugates |
| b2 -4ac = 0 |
One real (double root) |
| b2 - 4ac > 0 |
Two different real roots |
Warning!
Warning!
Danger Ahead!!
Be extremely
careful
in solving problems like this:
x2
= x
It is very tempting
to
divide both sides by x to get:
x = 1
This is incorrect,
because
you have completely eliminated one of the solutions. Never divide
both sides of an equation by a variable if it cancels from both sides!!
Correct way to do
the
problem is as follows:
x2
= x
x2
- x = 0
x(x - 1) = 0
Solutions are x = 0
and
x = 1
Honk!
Honk!
That about does it for this section! Onward and upward!
