Example 1-2: Quadratic Formula Example Problem
Given
24
3 + x =
(1-10)
x + 1
find x.
At first glance, this one doesn’t look like a quadratic equation, but as we begin isolating x, as we
always strive to do in solving for x, (hey, once we have x all by itself on the left side of the
equation, with no x on the right side of the equation, we have indeed solved for x—that’s what it
means to solve for x) we quickly find that it is a quadratic equation.
Whenever we have the unknown in the denominator of a fraction, the first step in isolating that
unknown is to multiply both sides of the equation by the denominator. In the case at hand, this
yields
(x + )
1 3
( + x) = 24
Multiplying through on the left we find
3x + 3
2
+ x + x = 24
At this point it is pretty clear that we are dealing with a quadratic equation so our goal becomes
getting it into the standard form of the quadratic equation, the form of equation 1-8, namely:
2
ax + bx + c = 0 . Combining the terms involving x on the left and rearranging we obtain
2
x + 4x + 3 = 24
Subtracting 24 from both sides yields
2
x + 4x − 21 = 0
which is indeed in the standard quadratic equation form. Now we just have to use inspection to
identify which values in our given equation are the a, b, and c that appear in the standard
quadratic equation (equation 1-8)
2
ax + bx + c = 0 . Although it is not written, the constant
multiplying the x2, in the case at hand, is just 1. So we have a = 1, b = 4, and c = −21.
Substituting these values into the quadratic formula (equation 1-9):
− b ± b2 − 4ac
x =
a
2
yields
− 4 ± 42 − 4 )(
1
(
− )
21
x =
2 )
1
(
which results in
x = 3 , x = 7
−
8