# modulus equation

• Apr 14th 2010, 11:06 AM
sigma1
modulus equation
am trying to solve for x in this equation

$\displaystyle x^2-\mid x \mid - 12 < 0$
$\displaystyle \mid x \mid <12-x^2$
$\displaystyle \mid x \mid^2 >{(12-x^2})^2$ squaring both sides

i end up with $\displaystyle x^4 - 24x^2+144=0$ can i get some helf solving for x. that if what i did was correct.
• Apr 14th 2010, 11:23 AM
running-gag
Hi

I suggest you to separate the 2 cases $\displaystyle x\leq 0$ and $\displaystyle x \geq 0$

In the first case you get $\displaystyle x^2 + x - 12 < 0$

$\displaystyle (x+4)(x-3) < 0$

Find the solutions and take the ones that are $\displaystyle \leq 0$

Do the same for the second case and finally join the 2 sets of solutions
• Apr 14th 2010, 11:28 AM
Quote:

Originally Posted by sigma1
am trying to solve for x in this equation

$\displaystyle x^2-\mid x \mid - 12 < 0$
$\displaystyle \mid x \mid <12-x^2$
$\displaystyle \mid x \mid^2 >{(12-x^2})^2$ squaring both sides

i end up with $\displaystyle x^4 - 24x^2+144=0$ can i get some helf solving for x. that if what i did was correct.

Hi sigma1,

you need c-ordination here,

you have made errors with polarity above.

$\displaystyle x^2-|x|-12<0$

If x is positive, this is

$\displaystyle x^2-x-12<0$

$\displaystyle (x-4)(x+3)<0$

this is negative if one factor is positive and the other negative.
If -3<x<4 the left factor is negative and the right one positive.
But remember, x must be positive here.
If x>4, they are both positive.
If x<-3 they are both negative.

If x is negative, this is

$\displaystyle x^2+x-12<0$

$\displaystyle (x-3)(x+4)<0$

This can be evaluated in the same way as above.