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!
