# Solve the inequality (x+3)(2x-3)<0

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(x+3)(2x-3) < 0

Since the result of the product is negative, then only one of the terms should be negative:

Then:

(x+3)<0 and (2x-3) > 0

==> x < -3 and x > 3/2 (impossible)

OR:

(x+3) > 0 and (2x-3) < 0

==> x > -3 and x < 3/2

==> -3 < x < 3/2

then the solution is:

**x belongs to the interval ( -3, 3/2) **

We have the inequality (x+3) (2x-3) <0.

Now (x+3) (2x-3) <0 when either of (x+3) and (2x-3) is less than 0 and the other is greater than 0.

First let's take (x+3)> 0 and (2x-3) <0

=> x> -3 and 2x <-3

=> x>-3 and x < -3/2

This gives x lies in the range where x > -3 and x< -3/2.

Next let's take x+3 <0 and 2x-3 > 0

=> x < -3 and 2x > -3

=> x < -3 and x > -3/2

This does not give valid values.

**Therefore for x: -3 < x < -3/2**

(x+3)(x-3) < 0

The left is a product of 2 linear epression. The product is less than zero. Or the product should be negative.

Therefore the values of both factors cannot be both positive together or both negative together in which case the product becomes positive.

So the possibility is x + 3 > 0 and x-3 < 0.

So If the value of lies between -3 and +3 , then x+3 < 0 and x -3 < 0

Therefore in order that (x+3)(x-3) < 0, the values of should bel within the interval of (-3 ,3) of the number line.

We'll conclude that a product is negative if the factors are of opposite sign.

There are 2 cases of study:

1) (2x-3) < 0

and

(x+3) > 0

We'll solve the first inequality. For this reason, we'll isolate 2x to the left side.

2x < 3

We'll divide by 2:

x < 3/2

We'll solve the 2nd inequality:

(x+3) > 0

We'll subtract 3 both sides:

x > -3

The common solution of the first system of inequalities is the interval (-3 , 3/2).

We'll solve the second case for the following system of inequalities:

2) (2x-3) > 0

and

(x+3) < 0

2x-3 > 0

We'll add 3 both sides:

2x > 3

x > 3/2

(x+3) < 0

x < -3

Since there is not a common interval to satisfy both inequalities, we don't have a solution for the 2nd case.

**So, the complete solution is the solution from the first system of inequalities, namely the interval (-3 , 3/2).**