Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to solve the inequality

$$|x^2-1|-2\ge 2x$$

but I am not sure where to start, because I have $x²$ in the absolute value part. $x²$ is always positive, and this is confusing for me. Can you please explain how to solve it?

share|improve this question
2  
Consider two cases, $x^2>1$ and $x^2\le1$. –  Macavity Feb 10 at 14:27
1  
$x^2$ is always positive, but $x^2 - 1$ (which is what is inside the absolute value sign) is not. –  Arthur Feb 10 at 14:29
    
So I have to treat it like $||x²|-1|$ –  depecheSoul Feb 10 at 14:30

3 Answers 3

up vote 2 down vote accepted

Factor $x^2 - 1$ and add $2$ on both sides to obtain

$|(x+1)(x-1)| \geq 2(x+1)$.

Now there are two cases. First, if $x \leq -1$, the RHS $2(x+1)$ is non-positive and hence the inequality is definitely fulfilled.

Second, let $x > -1$. But then, it follows that $|(x+1)(x-1)| = (x+1)|x-1|$, since $(x+1)$ is positive.

The inequality thus boils down to $(x+1)|x-1| \geq 2(x+1)$.

Eliminating the positive (by assumption of $x > -1$) factor $(x+1)$ then yields

$|x-1| \geq 2$.

This is either the case if $x-1 \geq 2$ ($\implies x \geq 3$), or $x-1 \leq -2$ ($\implies x\leq -1$). The second contradicts our assumption of $x > -1$, so $x\geq 3$ remains.

We conclude that the inequality is fulfilled if either (a) $x \leq -1$ or (b) $x \geq 3$.

share|improve this answer
    
How did you get x<=-1 and x>-1. Did you use x²-1=0; x²=1; x=+-1; Thanks. But the you have +1 also –  depecheSoul Feb 10 at 15:03
    
Do you mean the two cases? This follows directly from the right hand side - it will be negative for $x < -1$ and positive for $x \geq -1$. Hence, for $x < -1$, the inequality is automatically fulfilled. –  Martin Feb 10 at 15:17

$$|x^2 - 1| -2\ge 2x \iff |x^2 - 1| \ge 2x + 2 = 2(x + 1)$$ $$\iff |(x - 1)(x + 1)|\ge 2(x+1)$$

Now consider the cases $x^2\geq 1$ and $x^2 \lt 1$.

share|improve this answer
1  
does the case $x^2\ge1$ need to be treated as two sub-cases, $x\le1$ and $x\ge1$? –  TooTone Feb 10 at 14:44
    
For all (and only) those $x$ such that $-1\lt x \lt 1$, we have $x^2 \lt 1$ –  amWhy Feb 10 at 14:58

Another way this can be written:

One case is

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

and the other is

$$ (1 - x^2) \ - \ 2 \ \ge \ 2x \ \ \Rightarrow \ \ 0 \ \ge \ x^2 \ + \ 2x \ + \ 1 \ \ \Rightarrow \ \ 0 \ \ge \ (x + 1)^2 \ \ . $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.