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.


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}

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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