SOLVING RATIONAL INEQUALITIES WORKSHEET

Problem 1 :

Find all values of x for which

x3(x − 1)/(x − 2) > 0

Problem 2 : 

Find all values of x for which

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

Problem 3 : 

Find all values of x for which

(x2 - 4)/(x2 - 2x - 15)  0

Detailed Answer Key

Problem 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).

Problem 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).

Problem 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).

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