if you have the inequality:

$\displaystyle \frac{(x + 2)(x - 1)}{x + 1} < 0$

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.

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- Jan 24th 2010, 08:51 AMrenlokinequalities
if you have the inequality:

$\displaystyle \frac{(x + 2)(x - 1)}{x + 1} < 0$

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. - Jan 24th 2010, 09:10 AMJhevon
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 $\displaystyle x = -2,~-1, \text{ and } 1$ ($\displaystyle x \ne -1$ since the denominator cannot be zero, and $\displaystyle x \ne -2 \text{ or } 1$ 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:

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

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

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

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

Hence, your solution set is $\displaystyle (-\infty, -2) \cup (-1,1)$ - Jan 24th 2010, 09:34 AMArchie Meade
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,

$\displaystyle x<-2\ and\ -1<x<1$