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?
 A: 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 $\dfrac74<x\le2$.
A: 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 $f(x)<0$.
A: The radicand cannot be negative, so we must have $x\le 2.$ Also, the denominator is negative for $x\lt 1.$ Thus, you must consider this in the two cases when
(1) $x\lt 1,$ or when
(2) $1\lt x\le 2.$
What you've done (which depends crucially on your first step of squaring both sides) works only with the assumption in case (2), since then we have that $\text {LHS}\ge 0.$ Thus you can square both sides. In the first case, you cannot since then $\text {LHS}\lt 0$ whereas $\text {RHS}\gt 0.$ This is not true in general since, for example, the fact that $-3<1$ does not imply that $9=(-3)^2<1.$
Thus, in the first case you need to approach with a different method. In particular multiply both sides by the negative quantity $x-1$ to get $$3\sqrt{2-x}>2(x-1),$$ which is obviously true for any $x<1,$ since then $\text {LHS}\gt 0$ and $\text {RHS}\le 0.$ Thus the solution in this case is $$(-\infty,1),$$ as wanted.
