Find the solution set of $\frac{3\sqrt{2-x}}{x-1}<2$ Find the solution set of $\frac{3\sqrt{2-x}}{x-1}<2$
Start by squaring both sides
$$\frac{-4x^2-x+14}{(x-1)^2}<0$$
Factoring and multiplied both sides with -1
$$\frac{(4x-7)(x+2)}{(x-1)^2}>0$$
I got
$$(-\infty,-2)\cup \left(\frac{7}{4},\infty\right)$$
Since $x\leq2$ then
$$(-\infty,-2)\cup \left(\frac{7}{4},2\right]$$
But the answer should be $(-\infty,1)\cup \left(\frac{7}{4},2\right]$. Did I missed something?

Define $f(x)=\frac{3\sqrt{2-x}}{x-1}-2$.  Being continuous on its domain $(-\infty,2]\setminus\{1\}$,the function may change its sign only at its zero $7/4$  or at its singularity, namely at $1$. Now check the sign of $f$ in the corresponding intervals $(-\infty,1)$, $(1,7/4)$ and $(7/4,2]$; you want $f(x)<0$.
