Strategy to solve absolute value inequality

Question

I was wondering if there is any strategy to solve absolute value On both sides inequalities, for example,

$$| x^2 -3x + 2 | < | x + 2|$$

Thanks,

Eli

Answer 1

If you have one absolute value less than another, $|p|\lt|q|$, you can get an equivalent inequality by squaring both sides, to get $p^2\lt q^2$, after which you can write either $0\lt(q-p)(q+p)$ or $(p-q)(p+q)\lt0$, whichever seems nicer. In the OP's example, moving stuff to the left (which has the higher degree polynomial) gives

$$(x^2-4x)(x^2-2x+4)\lt0$$

From here it's a matter of factoring the two quadratics, listing their roots in increasing order, and then thinking about signs.

Answer 2

HINT:

Use for real $b,|b|=\begin{cases} +b &\mbox{if } b\ge0 \\ -b & \mbox{if } b<0\end{cases}$

and as $x^2-3x+2=(x-1)(x-2)$

Check the following cases:

$1)x=1,2,-2$

$2)x<-2$

$3)-2<x<1$

$4)1<x<2$

$5)x>2$

and also use $\displaystyle(x-a)(x-b)<0$ where $\displaystyle a<b;$

we can establish $\displaystyle a<x<b$