*
Section 1-6: Solving Quadratic Equations *
*Any equation in the
form*
*
ax*^{2} + 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) x*^{2}
+ 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
3x*^{2} + 10x - 8
= -11
*Put in correct
form
3x*^{2} + 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:*

*2x*^{2}
- 12x - 9 = 0
*Solution:*

*Move the constant to
the
other
side:
2x*^{2} - 12x = 9

*Divide by the
coefficient
of x*^{2}
x^{2} - 6x = 9/2

*Take half the
coefficient
of x and square: x*^{2}
- 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 ax*^{2}
+ bx + c = 0 then the roots of the equation are:
*Look
familiar?
It better!!*

*Solve the
following problem
by using the quadratic formula.*

*2x*^{2}
+ 5 = 3x
*
2x*^{2} - 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 b*^{2}
- 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 |

b^{2} - 4ac < 0 |
Two complex conjugates |

b^{2} -4ac = 0 |
One real (double root) |

b^{2} - 4ac > 0 |
Two different real roots |

*Warning!
Warning!
Danger Ahead!!*
*Be extremely
careful
in solving problems like this:*
*x*^{2}
= 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:*
*x*^{2}
= x
*x*^{2}
- 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!*