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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 x2 + 5 = 0  x2 = –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 x2 + 5 > 0  x2 > –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 x2 + 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:

(–5, 2) and (2, 3)

Note: If you'd factored the –2x – 10 and the 3 – x factors to convert them to the usual form for factors, you would have gotten the same zeroes, the same intervals, and the same final answer. The only difference would have been the inclusion of an additional factor in the factor table. First, you'd have factored:   Copyright © Elizabeth Stapel 2005-2011 All Rights Reserved

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

 2 > 0always x + 5 > 0x > –5 x – 3 > 0x > 3 x2 + 5 > 0x2 > –5always x – 2 > 0x > 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.

 Cite this article as: Stapel, Elizabeth. "Solving Rational Inequalities: Example." Purplemath. Available from     http://www.purplemath.com/modules/ineqrtnl2.htm. Accessed [Date] [Month] 2016

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