# inequalities

• January 24th 2010, 08:51 AM
renlok
inequalities
if you have the inequality:
$\frac{(x + 2)(x - 1)}{x + 1} < 0$

is there an easy way to find the values of x? as ive been just trying different numbers until it works which isnt really good method.
• January 24th 2010, 09:10 AM
Jhevon
Quote:

Originally Posted by renlok
if you have the inequality:
$\frac{(x + 2)(x - 1)}{x + 1} < 0$

is there an easy way to find the values of x? as ive been just trying different numbers until it works which isnt really good method.

yes, set the numerator and denominator equal to zero and solve for all the x-values you get. write these values on a number line, and test each interval (by plugging in any number in the interval) to see which numbers satisfy the inequality.

sigh, lets just do it.

Doing what I said, we get $x = -2,~-1, \text{ and } 1$ ( $x \ne -1$ since the denominator cannot be zero, and $x \ne -2 \text{ or } 1$ since that would make the expression equal to zero, which we don't want. so we will only have open intervals here).

our intervals are:

$(-\infty, -2)$ .......in this interval, test $x = -3$ in the original inequality. It works!

$(-2,-1)$ ........in this interval, test $x = -1.5$. it does NOT work.

$(-1,1)$ ........in this interval, test $x = 0$. It works!

and finally, $(1, \infty)$ .......in this interval, test $x = 2$. it does NOT work

Hence, your solution set is $(-\infty, -2) \cup (-1,1)$
• January 24th 2010, 09:34 AM
Alternatively, this is <0 if the numerator is + and the denominator is -
and vice versa.

The numerator is + if we have (-)(-) so x<-2.
This makes the denominator -

hence x<-2.

The numerator is also + for (+)(+), hence x>1.
This makes the denominator +, so x>1 does not obey the inequality.

The numerator is - for (-)(+) and (+)(-).

For x between -2 and 1 we have (+)(-) and the denominator is - for x<-1
hence the result of (-)/(-) is +,
but if the denominator is >-1, the result is -

hence -1<x<1 is a solution

The numerator is also - for (-)(+) but x<-2 gives (-)(-)

hence,

$x<-2\ and\ -1