Thread: Absolute value inequalities - I'm stuck...

1. Absolute value inequalities - I'm stuck...

Hi - I hoped someone maybe able to help me understand this question:
$\displaystyle absolutevalue\(\frac{(x+1)}{(x-3)(x-5)}) < \frac{1}{2}$

(apologies but I can't find the syntax for absolute value sign)

I know that I would need to square both sides in order to remove the modulus. But when rearranging I dont end up with a product so unsure how to proceed.

Thanks, F

2. Re: Absolute value inequalities - I'm stuck...

Originally Posted by FelixHelix
Hi - I hoped someone maybe able to help me understand this question:
$\displaystyle \left|\frac{(x+1)}{(x-3)(x-5)}\right| < \frac{1}{2}$

I know that I would need to square both sides in order to remove the modulus. But when rearranging I dont end up with a product so unsure how to proceed.
LaTeX Note: $$\left|\frac{(x+1)}{(x-3)(x-5)}\right| < \frac{1}{2}$$ gives
$\displaystyle \left|\frac{(x+1)}{(x-3)(x-5)}\right| < \frac{1}{2}$

If you do square.
$\displaystyle \frac{(x+1)^2}{(x-3)^2(x-5)^2} < \frac{1}{4}$.

$\displaystyle 4(x+1)^2<(x-3)^2(x-5)^2$.

Be careful: $\displaystyle x\ne 3~\&~x\ne 5$

3. Re: Absolute value inequalities - I'm stuck...

Originally Posted by Plato
LaTeX Note: $$\left|\frac{(x+1)}{(x-3)(x-5)}\right| < \frac{1}{2}$$ gives
$\displaystyle \left|\frac{(x+1)}{(x-3)(x-5)}\right| < \frac{1}{2}$

If you do square.
$\displaystyle \frac{(x+1)^2}{(x-3)^2(x-5)^2} < \frac{1}{4}$.

$\displaystyle 4(x+1)^2<(x-3)^2(x-5)^2$.

Be careful: $\displaystyle x\ne 3~\&~x\ne 5$
Thanks for the latex tip.

So with what you say in mind. If I draw a number line x cannot be equal to 5 or 3 but can be -1. And it will be always be positive between critical values because of the squaring of each parenthesis. Therefore the answers will be:

$\displaystyle x\le-1, -1< x <3, 3< x <5, x > 5$

Is this correct?

4. Re: Absolute value inequalities - I'm stuck...

Originally Posted by FelixHelix
$\displaystyle x\le-1, -1< x <3, 3< x <5, x > 5$
Is this correct?
That nowhere close to correct!
For one, $\displaystyle x=2$ is not a solution.
Is $\displaystyle x=6~?$