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Given inequality $(x - 2)\sqrt{x^2 + 1} > x^2 + 2$, find it's solution as intervals. And I have problem solving it.

So at first, both $\sqrt{x^2 + 1} > 0$ and $x^2 + 2 > 0$. That means, that we have to look for values of $(x - 2)$. The equality is true for $x \geq 2$, so we are solving for it. After taking all terms to 2nd power, we get $x(4x^2 - x + 4) < 0$, which is obviously true for $x < 0$. Combining all of that, solution $S = \emptyset$.

The question, that is bothering me is, am I required find solutions for $x < 2$? (And then of course applying change of inequality sign).

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    $\begingroup$ If $x < 2$ the inequality is obviously false: What sign does each part have? $\endgroup$ – user296602 Jun 6 '16 at 20:34
  • $\begingroup$ Left side is minus, right side is plus - then it is obviously false. But shouldn't I reverse inequality sign in that case? $\endgroup$ – Accelerate to the Infinity Jun 6 '16 at 20:36
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    $\begingroup$ No. The given inequality goes one way; why would you reverse it? That's something completely different. $\endgroup$ – user296602 Jun 6 '16 at 20:38
  • $\begingroup$ Thanks for clarification! May I ask about cases, when I shall reverse the sign? $\endgroup$ – Accelerate to the Infinity Jun 6 '16 at 20:39
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    $\begingroup$ You don't get to choose the inequality, it is given to you. You only get to choose values of $x$ to propose as solutions. Any value of $x$ that is less than $2$ produces a statement that is false, so no such value can be a solution (you shouldn't propose it as a member of the solution set). $\endgroup$ – MPW Jun 6 '16 at 20:40
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As you noted, $\sqrt{x^2 + 1} > 0$ and $x^2 + 2>0$ for all $x\in \mathbb{R}$.

The reason why there is no solution $x\leq 2$ is that if there was, then $0\geq (x-2)$, which means $0\geq (x-2)\sqrt{x^2 + 1}$. Hence $$0\geq(x - 2)\sqrt{x^2 + 1} > x^2 + 2>0$$ which is impossible.

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    $\begingroup$ I don't follow this at all. I agree that there are no values of $x\le2$ for which the given equality is true. But the statement that $x-2>\frac{x^2+2}{\sqrt{x^2+1}}$ is false for all $x$. $\endgroup$ – almagest Jun 6 '16 at 20:51
  • $\begingroup$ Since we were given the inequality $(x - 2)\sqrt{x^2 + 1} > x^2 + 2$, I divided both sides by the positive value $\sqrt{x^2 + 1}$. It had no solutions to begin with, so that is why its false. I maybe should have been more explicit. $\endgroup$ – M47145 Jun 6 '16 at 20:54
  • $\begingroup$ All I am saying is that it is fairly confusing read on its own. It would be clearer for example if you said $\frac{x^2+2}{\sqrt{x^2+1}}>0>x-2$. $\endgroup$ – almagest Jun 6 '16 at 20:56
  • $\begingroup$ @almagest Assuming $x<2$? $\endgroup$ – Fly by Night Jun 6 '16 at 20:57
  • $\begingroup$ @almagest I edited it. I hope it is more clear now. $\endgroup$ – M47145 Jun 6 '16 at 21:02

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