# Finding solution for irrational inequality

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?

• 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. $1 < x < 9$. – Crostul Jun 7 '16 at 22:23
• This is a rare example of a math question whose title could also be a news headline. – Akiva Weinberger Jun 7 '16 at 23:12

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 $x \in (1,9)$
• 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 $- \times -$). – Accelerate to the Infinity Jun 7 '16 at 22:42