The set of values of x for which $x^{2}-3\left | x \right |+2< 0$ is given by?

Question

So i could come up with $$(x-1)(x-2)<0$$

and whatever will be the solution it will also extend to the negative value of the $x's$ because the sign does not matter in this equation.

However i don't know how to proceed, Any help will be appreciated!

Answer 1

Set $t=|x|$, so the inequality becomes $t^2-3t+2<0$, which is satisfied for $1<t<2$. Hence the original one is satisfied for $1<|x|<2$, which is the same as $$ -2<x<-1\qquad\text{or}\qquad 1<x<2 $$

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Just to make a step forward, if you have to solve $x^2-3|x|+2>0$, you can do the same and get $$ t<1 \qquad\text{or}\qquad t>2 $$ Now the inequalities become $$ |x|<1 \qquad\text{or}\qquad |x|>2 $$ which give the full set $$ x<-2 \qquad\text{or}\qquad -1<x<1 \qquad\text{or}\qquad x>2 $$

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On the other hand, an inequality such as $$ x^2-|x|-2<0 $$ would give $-1<t<2$ which is the same as $|x|<2$. Why?

Answer 2

If $x \geq 0$ then the condition becomes $x^{2} - 3x + 2 < 0$; if $x < 0$ then the condition becomes $x^{2} -3(-x) + 2 = x^{2} + 3x + 2 < 0$.

Answer 3

you will get for $$x\geq 0$$ $$x^2-3x+2$$ factorizing this product we get $$(x-1)(x-2)$$ for $$x<0$$ we get $$x^2+3x+2$$ and factorizing this we get the product $$(x+2)(x+1)$$