<s>[INST] 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?
 [/INST] For $\dfrac{3\sqrt{2-x}}{x-1}$ to be defined, $x\le2$ and $x\ne1$.
If $x<1,$ then the expression is negative (i.e., $<0$), so of course it is $< 2$.
If $x>1,$ then, as you showed, the inequality holds when $x>\dfrac74$.
Therefore, the solution set is $x<1$ or $\dfrac74<x\le2$.
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