Solving rational inequalities is very similar to solving polynomial inequalities.
But because rational expressions have denominators (and therefore may have places where they're not defined), you have to be a little more careful in finding your solutions.
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To solve a rational inequality, use these steps:
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Let's see how these instructions work in practice:
They've already put this inequality into (one rational expression) with (zero) on the other side. So I can start with factoring everything:
This polynomial fraction will be zero wherever the numerator is zero, so I'll set the numerator equal to zero and solve:
(x + 2)(x + 1) = 0
x + 2 = 0 or x + 1 = 0
x = −2 or x = −1
The fraction will be undefined wherever the denominator is zero, so I'll set the denominator equal to zero and solve:
(x + 4)(x − 4) = 0
x + 4 = 0 or x − 4 = 0
x = −4 or x = 4
These four values, −4, −2, −1, and +4, divide the number line into five intervals, namely:
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(−∞, −4)
(−4, −2)
(−2, −1)
(−1, 4)
(4, +∞)
I could use "test points" to find the solution to the inequality, by picking an x-value in each interval, plugging it into the original rational expression, simplifying to get a numerical answer, and then checking the sign, but that process gets long and annoying (and is prone to errors), so I'll use the easier and faster factor-table method instead.
My factor table looks like this:
My table has one row for each factor, a row for the number line, and a row for the rational expression. Each row is split into columns, with each column corresponding to one of the intervals on the number line.
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The sign of the overall rational expression is a result of the signs of its various factors, so I need to find where each factor is positive:
x + 4 > 0 for x > −4
x + 2 > 0 for x > −2
x + 1 > 0 for x > −1
x − 4 > 0 for x > 4
Now I can put "plus" signs on the intervals in each row where that row's factor is positive:
Wherever a factor isn't positive, it's negative, so I'll put "minus" signs in the remaining entries of each row:
I know that the product of an even number of "minus" signs is a plus; the product of an odd number of "minus" signs is a minus. So, by multiplying the signs down the columns (or just counting up the minusses), I get the overall sign of the rational expression on each interval:
Then the rational is positive on the intervals (−∞, −4), (−2, −1), and (4, +∞).
Looking back at the original exercise, this is an "or equal to" inequality, so I need to consider the interval endpoints, too.
If this were a polynomial inequality, I could just throw all the interval endpoints into the solution, and I'd be done. For rational expressions, though, I have to be careful not to include any x-values that would cause division by zero.
The intervals' endpoints are −4, −2, −1, and 4. I can include −2 and −1 in the solution, because they just make the expression equal to zero by making the numerator zero. But plugging −4 or 4 into the rational expression would cause division by zero, making the rational expression undefined, so I can't include these values in the solution.
Then my full solution is:
I wrote my solution above in "interval" notation. If you have to write your solution in "inequality" notation, it would look like this:
x < −4, −2 ≤ x ≤ −1, and x > 4
Don't forget: "Infinity" is not a "number" in the way that, say, "2" is. "Infinity" cannot be "included" in your solution, so never draw a square bracket next to an "infinity" "endpoint".
URL: https://www.purplemath.com/modules/ineqrtnl.htm
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