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Determine the values of $x$ satisfying the equality $|3x^2-5|-|2x^2+3|=|x^2-8|$

I can see that it is written in the form $|x|-|y|=|x-y|$ which is true when $x$ and $y$ are of different signs so $xy\le0$ but that does not give the correct answer.

Please tell what am i doing wrong and what is the correct method to solve these kind of problems.

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You are wrong about when the equality $|x|-|y|=|x-y|$ holds. It holds when (and only when) of of these conditions holds:

  • $x\geqslant y\geqslant0$;
  • $x\leqslant y\leqslant0$.

I shall now use this to solve your problem. The equality holds if and only if$$3x^2-5\geqslant2x^2+3\geqslant0\text{ or }3x^2-5\leqslant2x^2+3\leqslant0.$$The secons possibility cannot take place, of course. On the other hand\begin{align}3x^2-5\geqslant2x^2+3&\iff x^2\geqslant8\\&\iff x\geqslant\sqrt8\vee x\leqslant-\sqrt8.\end{align}

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  • $\begingroup$ so $x$ and $y$ should be of same sign, which implies $xy>0$ $\endgroup$ – Gem Aug 16 '17 at 14:52
  • $\begingroup$ @Gem Yes, of course. Otherwise $|x-y|>\bigl||x|-|y|\bigr|\geqslant|x|-|y|$. $\endgroup$ – José Carlos Santos Aug 16 '17 at 14:55
  • $\begingroup$ but xy>0 is not giving the correct answer $\endgroup$ – Gem Aug 16 '17 at 14:56
  • $\begingroup$ @Gem The solution is $x\geqslant\sqrt8\vee x\leqslant-\sqrt8$. Do you agree? $\endgroup$ – José Carlos Santos Aug 16 '17 at 15:03
  • $\begingroup$ yes but how do you get it by solving $(3x^2−5)*(2x^2+3)>0$ $\endgroup$ – Gem Aug 16 '17 at 15:05

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