# SOLVING RATIONAL INEQUALITIES

Solving Rational Inequalities :

In this section, we will learn, how to solve rational inequalities.

## Solving Rational Inequalities - Examples

Example 1 :

Find all values of x for which

x3(x − 1)/(x − 2) > 0

Solution :

x3(x − 1)/(x − 2) > 0

Thus, x3(x − 1) and (x − 2) are both positive or both negative.

So let us find out the signs of x3(x − 1) and x - 2 as follows

 Values of x x3 x−1 x-2 x3(x−1)/(x−2) x < 0 - - - - 0 < x < 1 + - - + 1 < x < 2 + + - - x > 2 + + + +

The values from the intervals (0, 1) and (1, 2) satisfies the above rational inequality.

Hence, the solution is (0, 1) U (1, 2).

Example 2 :

Find all values of x for which

(2x - 3)/(x - 2) (x - 4) < 0

Solution :

(2x - 3)/(x - 2) (x - 4) < 0

Let f(x)  =  (2x - 3)/(x - 2) (x - 4)

Thus, (2x - 3) and (x−2) (x - 4) are both positive or both negative.

So let us find out the signs of (2x - 3) and (x−2) (x-4)  as follows

 Values of x 2x-3 (x-2)(x-4) f(x) x < 3/2 - + - 3/2 < x < 2 + + + 2 < x < 4 + - - x > 4 + + +

The values from the intervals (-, 3/2) and (2, 4) satisfies the above rational inequality.

Hence, the solution is (-, 3/2) (2, 4).

Example 3 :

Find all values of x for which

(x2 - 4)/(x2 - 2x - 15)  0

Solution :

(x2 - 4)/(x2 - 2x - 15)  0

Let us factorize both numerator and denominator.

(x2 - 4)/(x - 5) (x + 3)  0

Let f(x)  =  (x + 2) (x - 2)/(x - 5) (x + 3)

Thus, (x + 2), (x - 2), (x - 5) and (x + 3) are both positive or both negative.

So let us find out the signs of (x + 2), (x - 2), (x - 5) and (x + 3) as follows

 Values of x (x+2) (x-2) (x-5) (x+3) f(x) x < -3 + + + -3 < x < -2 + - - -2 < x < 2 - - + 2 < x < 5 + - - x > 5 + + +

The values from the intervals (-3, -2] and [2, 5) satisfies the above rational inequality.

Hence the solution is (-3, -2] and [2, 5).

After having gone through the stuff given above, we hope that the students would have understood, how to solve rational inequalities.

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