Say you have $\frac{2x^2+3x-2}{2x^2-3x-2}>0$
How do you find the range of solutions? Can you just bring the denominator onto the right hand side or does that lose solutions? Never been taught how to approach these kind of inequalities.
BTW: This is not for homework or anything of a similar kind. It's just a problem that I came across in my free time.
On
Another way of doing this kind of question is to multiply both sides by the denominator squared which, being non-negative, preserves the inequality.
So you could go straight to
$$(2x^2+3x-2)(2x^2-3x-2)>0$$ $$\implies(2x-1)(x+2)(2x+1)(x-2)>0$$
You then just have to sketch a quick quartic graph showing the four roots and read off the solution sets.
No, you cannot bring the denominator to the right. That's equivalent to multiply the fraction by $2x^2-3x-2$, but then the inequality should change when this number is smaller then $0$.
What you can do is to determine those $x\in\Bbb R$ in which $2x^2+3x-2$ and $2x^2-3x-2$ have the same sign. After all $\frac ab>0$ if and only if $a$ and $b$ aro both positive or both negative.