[ (x-1)²-(x+5) ] / (x+5) < 0

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- Jun 4th 2011, 06:30 PMsilvercatsHow to solve this inequality ?
[ (x-1)²-(x+5) ] / (x+5) < 0

- Jun 4th 2011, 06:54 PMProve It
$\displaystyle \displaystyle \frac{(x - 1)^2 - (x + 5)}{x + 5} < 0$

First note that $\displaystyle \displaystyle x\neq -5$ since this would give a zero denominator.

Now you need to test two separate cases, one where the denominator is negative, and one where the denominator is positive (because multiplying or dividing by a negative number changes the inequality sign).

Case 1: $\displaystyle \displaystyle x + 5 < 0 \implies x < -5$

$\displaystyle \displaystyle \begin{align*} \frac{(x - 1)^2 - (x + 5)}{x + 5} &< 0 \\ (x - 1)^2 - (x + 5) &> 0 \\ x^2 - 2x + 1 - x - 5 &> 0\\ x^2 - 3x - 4 &> 0 \\ x^2 - 3x + \left(-\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right)^2 - 4 &> 0 \\ \left(x - \frac{3}{2}\right)^2 - \frac{9}{4} - \frac{16}{4} &> 0 \\ \left(x - \frac{3}{2}\right)^2 - \frac{25}{4} &> 0 \\ \left(x - \frac{3}{2}\right)^2 &> \frac{25}{4} \\ \left|x - \frac{3}{2}\right| &> \frac{5}{2}\end{align*}$

So $\displaystyle \displaystyle x - \frac{3}{2} < -\frac{5}{2} \implies x < -1$ or $\displaystyle \displaystyle x - \frac{3}{2} > \frac{5}{2} \implies x > 4$. Since in this case, we already said that $\displaystyle \displaystyle x < -5$, that means the solution set for this case is $\displaystyle \displaystyle x < -5$.

Now you try Case 2, where $\displaystyle \displaystyle x + 5 > 0 \implies x > -5$. - Jun 5th 2011, 02:42 AMArchie Meade
As an alternative,

you either have "positive numerator" and "negative denominator"

or "negative numerator" and "positive denominator".

Those are the only ways the fraction can be negative.

$\displaystyle x<-5\Rightarrow\ (x-1)^2-(x+5)>0\Rightarrow\ x^2-2x+1-x-5>0$

$\displaystyle \Rightarrow\ x^2-3x-4>0\Rightarrow\ (x-4)(x+1)>0$

Both of these factors must be positive or both must be negative.

Both are positive for $\displaystyle x>4$ but then we cannot have $\displaystyle x>4$ and $\displaystyle x<-5$ at the same time.

Both are negative for $\displaystyle x<-1$ and for $\displaystyle x<-1$ and $\displaystyle x<-5$ at the same time, then $\displaystyle x<-5$

$\displaystyle x>-5\Rightarrow\ (x-1)^2-(x+5)<0\Rightarrow\ (x-4)(x+1)<0$

These factors must have opposite sign.

For the first positive and the second negative...

$\displaystyle x>4$ and $\displaystyle x<-1$ and the second conflicts with the first.

For the first negative and the second positive...

$\displaystyle x<4$ and $\displaystyle x>-1$ which is consistent with $\displaystyle x>-5$ - Jun 5th 2011, 06:57 AMsilvercats
Thanks you both!But I thought we can't multiply both sides by (x+5) ,If it is a inequality :o .That is what I've heard before

- Jun 5th 2011, 07:06 AMskeeter
find critical values ...

look for the values of x that make the numerator 0 ... x = 4 , x = -1

look for the values of x that make the denominator 0 ... x = -5

these three x-values break the number line into four regions ... since the rational expression is continuous in each region, pick any single value of x in each region and determine if that x-value makes the expression positive or negative.

the region(s) where the expression is negative is your solution set ...

x < -5 , -1 < x < 4 - Jun 5th 2011, 08:22 AMArchie Meade
You need to bear in mind that 2>1 but -2<-1.

This scenario can occur if you multiply or divide both sides of an inequality by a negative number.

Hence, for a variable (such as x) in the denomimator, you must account for the possibility

of the denominator being negative.

Then, if you multiply both sides by the negative denominator, you must reverse the inequality sign.

If the denominator is positive, no reversal is necessary.

But you have to account for these cases.

The method I gave avoids this. - Jun 5th 2011, 08:38 AMProve It
- Jun 5th 2011, 08:49 AMsilvercats
oh I see....... I can understand now.Thanks Archie Meade,Prove It and skeeter for the great answers