1
$\begingroup$

I have given an inequality: $\frac{x-2\sqrt{x} - 3}{x + \sqrt{x} - 2} < 0$.

Now trying to solve it, I'm doing substitution for $a = \sqrt{x}$ and then solving irrational equations as quadratics $a^2 - 2a - 3 = 0$ and $a^2 + a - 2 = 0$ to obtain roots. By solving those equations, I have the following: $$\frac{(\sqrt{x} - 3)(\sqrt{x} + 1)}{(\sqrt{x} + 2)(\sqrt{x} - 1)} < 0.$$

How can I now determine solution of the inequality?

$\endgroup$
2
  • 1
    $\begingroup$ Recall that $\sqrt{x} \ge 0$, so that $\sqrt{x}+1 \ge 1 >0$ and $\sqrt{x}+2 \ge 2 >0$. So you are left with $$\frac{\sqrt{x}-3}{\sqrt{x}-1} <0$$, whose solution is $1<\sqrt{x}<3$, i.e. $1 < x < 9$. $\endgroup$
    – Crostul
    Jun 7, 2016 at 22:23
  • 2
    $\begingroup$ This is a rare example of a math question whose title could also be a news headline. $\endgroup$ Jun 7, 2016 at 23:12

1 Answer 1

2
$\begingroup$

Let $a=\sqrt x$, as you've done, and consider $g(a) := (a^2 - 2a - 3)(a^2 + a - 2) = (a-3)(a+1)(a+2)(a-1)$ defined for non-negative $a$. One can verify (with, say, a sign chart) that $g(a) < 0 \Longleftrightarrow a \in (1,3)$.

Thus, noting that $x \mapsto \sqrt x$ is continuous and increasing, the solution is $x \in (1,9)$

$\endgroup$
2
  • $\begingroup$ Can I use the following reasoning? If $a > 0$, then terms $(a + 1)$ and $(a + 2)$ are always strictly positive. It leaves us then with terms $(a - 3)$ and $(a - 1)$. Term $(a - 3)$ is negative for all $a < 3$. The term $(a - 1)$ should be positive in that case, which is only when $a > 1$ (because otherwise it would be $- \times -$). $\endgroup$ Jun 7, 2016 at 22:42
  • 1
    $\begingroup$ @AcceleratetotheInfinity Yeah, that seems fine $\endgroup$ Jun 7, 2016 at 22:44

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.