# How to solve this inequality ?

• June 4th 2011, 07:30 PM
silvercats
How to solve this inequality ?
[ (x-1)²-(x+5) ] / (x+5) < 0
• June 4th 2011, 07:54 PM
Prove It
$\displaystyle \frac{(x - 1)^2 - (x + 5)}{x + 5} < 0$

First note that $\displaystyle x\neq -5$ 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: $\displaystyle x + 5 < 0 \implies x < -5$

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

Now you try Case 2, where $\displaystyle x + 5 > 0 \implies x > -5$.
• June 5th 2011, 03:42 AM
Quote:

Originally Posted by silvercats
[ (x-1)²-(x+5) ] / (x+5) < 0

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.

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

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

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

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

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

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

These factors must have opposite sign.
For the first positive and the second negative...
$x>4$ and $x<-1$ and the second conflicts with the first.

For the first negative and the second positive...
$x<4$ and $x>-1$ which is consistent with $x>-5$
• June 5th 2011, 07:57 AM
silvercats
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
• June 5th 2011, 08:06 AM
skeeter
Quote:

Originally Posted by silvercats
[ (x-1)²-(x+5) ] / (x+5) < 0

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
• June 5th 2011, 09:22 AM
Quote:

Originally Posted by silvercats
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

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.
• June 5th 2011, 09:38 AM
Prove It
Quote:

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.

So did the method I gave - i.e. to treat the solution process as two separate cases, one where the denominator is negative, the other where the denominator is positive.
• June 5th 2011, 09:49 AM
silvercats
oh I see....... I can understand now.Thanks Archie Meade,Prove It and skeeter for the great answers