Mat 108 Intermediate Algebra
Unit 4
Part 4
Quadratic Equations and the Quadratic Formula
Apparently, a mathematician (maybe Bombelli in the 16
th
century) tired of solving
quadratic equations by
completing the square
and decided to shortcut this repetitive
process.
He followed the same process (
completing the square
) used for centuries but on the
general quadratic
2
0
ax
bx
c
+
+ =
with coefficients of
a
,
b
, and
c
rather than one specific
example after another.
The result of his work led to the following theorem:
If a quadratic equation in one variable is in
standard form
(
2
0
ax
bx
c
+
+ =
) and the
coefficients (
a
,
b
, and
c
) are any
complex numbers
, then the solutions are as follows:
2
2
2
4
4
4
2
2
2
b
b
ac
b
b
ac
b
b
ac
or
or
a
a
a

+





ﾱ

Further more, if we call these solutions
1
r
and
2
r
, then
1
2
1
2
b
c
r
r
and r
r
a
a
+
=
=
g
.
This second part of the theorem is often called the
root test
and can be used to
check
solutions without substituting them in the original problem.
Let’s try Example 1:
Solve
2
3
2
3
x
x
=
+
using the
quadratic formula
.
First, we must put the quadratic equation in
standard form
:
2
2
3
2
3
2
3
2
3
3
2
3
0
x
x
x
x
x
x
=
+






=
Next, we have to identify
a
,
b
, and
c
.
For this example (now that it’s in
standard form
), we see that
3;
2;
3
a
b
c
=
=
=
and
we substitute these into the formula:
(
)
(
)
(
) (
)
(
)
2
2
4
2
2
2
4 3
3
2 3
b
b
ac
a

ﾱ



ﾱ



1
Simplifying, we have
(
)
2
4
36
2
40
2
2 10
40
2 10,
6
6
6
or
simplifying the
as
we have
ﾱ
+
ﾱ
ﾱ
The last expression is reducible (there’s a GCF of “2” in all three terms):
1
10
3
ﾱ
.
We believe that the solution set for this quadratic equation is
1
10
3
�
�
ﾱ
�
�
�
�
�
�
�
Using the second part of the theorem, we can check these answers without actually
substituting them into the original equation.
Let’s label our answers:
1
2
1
10
1
10
3
3
r
and r
+

=
=
.
If we can show that
1
2
1
2
b
c
r
r
and r
r
a
a
+
=
=
g
(i.e. both are true), we will
know
that our
answers are correct.
Let’s try showing that
1
2
b
r
r
a
+
=
(remember that
3;
2;
3
a
b
c
=
=
=
):
1
2
1
10
1
10
2
3
3
3
r
r
�
� �
�
+

+
=
+
=
�
� �
�
�
� �
�
�
� �
�
(They have the same denominators and the two radicals would cancel.)
Is this result (the
2
3
), the same as the result if we evaluate
b
a

?
Let’s see:
(
)
(
)
2
2
3
3
b
a


ﾱ

ﾱ
Yes it is.