# Solve the inequality (2x-1)(x+2)<0

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### 4 Answers

(2x-1) (x+2) < 0

since the value is negative then we have two cases:

case (1):

2x-1 <0 AND x+ 2 > 0

x < 1/2 AND x >-2

==> -2 < x < 1/2

==> x belongs to the interval (-2, 1/2)

Case 2:

2x-1 > 0 AND x+2 < 0

x > 1/2 AND x < -2

==> x = (1/2, inf) + (-inf, -2) = empty set.

Then the solution is :

**x belongs to ( -2, 1/2)**

The product of two terms is negative if either of the terms is negative and the other is positive.

For (2x-1)(x+2)<0

- 2x-1 < 0 and x+2 >0

=> 2x < 1 and x > -2

=> x < 1/2 and x > -2

So x lies between -2 and 1/2

- 2x - 1> 0 and x+2 <0

=> 2x > 1 and x < -2

=> x > 1/2 and x< 2

This is not possible.

**Therefore the solution is x lies between -2 and 1/2.**

(2x-1)(x+2)<0.

To solve the inequality.

Solution:

First we consider the roots of the equation; (x+2)(2x-1) = 0.

x = -2 and x= 1/2/

The product on the left is less than zero if only one of the factors is negative and the other is positive .

Product is positive if both of the factors are positive or both of the factors are negative.

So the product is negative only if the value of x is between the roots -2 and 1/2. Or x belongs to the interval (-2 , 1/2)

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

There are 2 caes of study:

1) (2x-1) < 0

and

(x+2) > 0

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

2x < 1

We'll divide by 2:

x < 1/2

We'll solve the 2nd inequality:

(x+2) > 0

We'll subtract 2 both sides:

x > -2

The common solution of the first system of inequalities is the interval (-2 , 1/2).

We'll solve the second systemof inequalities:

2) (2x-1) > 0

and

(x+2) < 0

2x-1 > 0

We'll add 1 both sides:

2x > 1

x > 1/2

(x+2) < 0

x < -2

Since we don't have a common interval to satisy 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 (-2 , 1/2).**