Find the solution set of $\frac{3\sqrt{2-x}}{x-1}<2$

Question

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?

From $-3<2$ it doesn't follow that $(-3)^2<2^2$. – Michael Hoppe Jun 18 '19 at 13:08 — Michael Hoppe, Jun 18 '19 at 13:08
seems similar to this question – J. W. Tanner Jun 18 '19 at 13:08 — J. W. Tanner, Jun 18 '19 at 13:08

score 4 · Accepted Answer · answered Jun 18 '19 at 13:22

4

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 $\dfrac74<x\le2$.

answered Jun 18 '19 at 13:22

J. W. Tanner

60,406

2

Sometimes the simplest answers are the best +1 – 1123581321 Jun 19 '19 at 09:27

score 0 · Answer 2 · edited Jun 19 '19 at 13:39

0

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 $f(x)<0$.

edited Jun 19 '19 at 13:39

J. W. Tanner

60,406

answered Jun 18 '19 at 13:13

Michael Hoppe

18,103
3
32
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The downvote was what for? – Michael Hoppe Jun 18 '19 at 15:43
I didn't downvote, but $-2$ is not a zero of $f(x)$ – J. W. Tanner Jun 18 '19 at 23:23
Holy macaron, how dumb of me. Thanks for pointing that out. Corrected. – Michael Hoppe Jun 19 '19 at 08:22

score 0 · Answer 3 · answered Jun 19 '19 at 10:27

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.

Find the solution set of $\frac{3\sqrt{2-x}}{x-1}<2$

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