# inequality confucion

• Jan 15th 2009, 11:43 PM
inequality confucion
Solve this inequality .

$\displaystyle \frac{6}{|x|+1}<|x|$

squarring both sides , i got

$\displaystyle \frac{36}{x^2+2|x|+1}<x^2$ , rearranging them

$\displaystyle \frac{36}{x^2+2|x|+1}-\frac{x^2(x^2+2|x|+1)}{x^2+2|x|+1}<0$

$\displaystyle \frac{36-x^4-2x^2|x|-x^2}{x^2+2|x|+1}<0$ , till here am i correct , and i don see a way to solve it from here .
• Jan 15th 2009, 11:53 PM
Isomorphism
Quote:

Solve this inequality .

$\displaystyle \frac{6}{|x|+1}<|x|$

squarring both sides , i got

$\displaystyle \frac{36}{x^2+2|x|+1}<x^2$ , rearranging them

Squaring unnecessarily complicates this simple problem. Instead take the following route:

$\displaystyle \frac{6}{|x|+1}<|x|$

$\displaystyle \frac{6}{|x|+1} - |x| < 0$

$\displaystyle \frac{6 - |x|^2 - |x|}{|x|+1} < 0$

But $\displaystyle |x| + 1 > 0$, thus we can multiply it on both sides without affecting the inequality. Thus

$\displaystyle 6 - |x|^2 - |x| < 0$

And so on....
• Jan 16th 2009, 03:34 AM
Re :
Quote:

Originally Posted by Isomorphism
Squaring unnecessarily complicates this simple problem. Instead take the following route:

$\displaystyle \frac{6}{|x|+1}<|x|$

$\displaystyle \frac{6}{|x|+1} - |x| < 0$

$\displaystyle \frac{6 - |x|^2 - |x|}{|x|+1} < 0$

But $\displaystyle |x| + 1 > 0$, thus we can multiply it on both sides without affecting the inequality. Thus

$\displaystyle 6 - |x|^2 - |x| < 0$

And so on....

Yeah , i make things complicated for myself .
From there , i got x>3 and x<-3 .... is it correct .
• Jan 16th 2009, 07:25 AM
earboth
Quote:

Yeah , i make things complicated for myself .
From there , i got x>3 and x<-3 .... is it correct .

Unfortunately no.

From $\displaystyle 6-|x|^2-|x|<0$ ......you'll get:

$\displaystyle |x|^2+|x|-6>0~\implies~(|x|+3)(|x|-2)>0$

A product of 2 factors is positive if both factors are positive or both factors are negative. Since $\displaystyle |x|+3>0$ is true for all x, the second case is impossible. Therefore (keep in mind that $\displaystyle |x|+3>0$ is true for all x):

$\displaystyle |x|-2>0~\implies~|x|>2$

Now use the definition of the absolute value:

$\displaystyle |x|>2~\iff~x>2~\vee~-x>2~\iff~\boxed{x<-2~\vee~x>2}$

By the way:
Quote:

From there , i got x>3 and x<-3
You'll never find a number which is greater than 3 and simultaneously is smaller than -3.