Solving
Rational Inequalities: Example (page
2 of 2)

In the previous example,
the sign of the rational expression alternated with the intervals.
Don't assume that this will always be true.

Solve the following:

This is already factored
for me, so I don't have to bother with that. I'll go straight to finding
the zeroes (from the numerator) and the undefined points (from the denominator):

–2x
– 10
= 0 –2x
= 10 x
= –5

3
– x
= 0 3
= x

x^{2}
+ 5 = 0 x^{2}
= –5
no
solution

x
– 2
= 0 x
= 2

So the number line is
split into the intervals (–infinity,
–5),
(–5,
2), (2, 3), and (3,
+infinity). Now I'll
find where each factor is positive:

–2x
– 10
> 0 –2x
> 10 x
< –5

3
– x
> 0 3
> x x
< 3

x^{2}
+ 5 > 0
x^{2}
> –5
always
true

x
– 2
> 0 x
> 2

The negative factor,
–2x
– 10, and the
"backwards" factor, 3
– x, gave
me "backwards" inequalities, so the factor table looks like
this:

(The "x
– 2"
factor is listed twice, because that factor occurs twice. If that factor
had been cubed, it would have been listed in the table three times.
The x^{2}
+ 5 factor has
all "plus" signs in its row, because this factor is never
zero or negative.)

Looking at the signs in
the bottom row, I see that the rational expression is negative on the
intervals (–5,
2) and (2,
3). Since this problem
is not an "or equal to" inequality, I don't need to consider
the endpoints; I know they don't belong in the solution. But I do need
to remember that x
= 2 is not part
of the solution, and I must resist the impulse to join these two intervals
together (by throwing in x
= 2) to get an incorrect
solution of (–5,
3).

Since I can't include x
= 2, the solution is
two separate intervals:

Then you'd have found the
endpoints and the signs on each interval:

2 >
0 always

x
+ 5 > 0 x > –5

x
– 3
> 0 x > 3

x^{2}
+ 5 > 0 x^{2} > –5 always

x
– 2
> 0 x > 2

And then you'd have filled
out your (only slightly longer) factor table, and would then have read
off the solution from the bottom row:

The solution would still
have been the same two intervals: (–5,
2) and (2,
3)

When working these problems,
remember to be careful of constant factors (like "2")
and backwards factors (like "3
– x").
And make sure to be careful about which endpoints you include for "or
equal to" inequalities. But as long as you are methodical in factoring,
in finding the zeros and the undefined points, and in finding the signs
of each factor on each interval, you should consistently get the right
answers.