Question

Solve the irrational inequality:
$\frac{3}{\sqrt{2-x}}-\sqrt{2-x}\le 2$

• A. $(-5,3)$
• B. $[-7,1]$
• C. $(-\infty ,-1)$
• D. $(-\infty ,1]$

Answer 1

The correct option is D $(-\infty ,1]$

Given $\frac{3}{\sqrt{2-x}}-\sqrt{2-x}\le 2$ will be true
If $2-x>0\Rightarrow x<2\cdots \left(1\right)$

Now
$\frac{3}{\sqrt{2-x}}-\sqrt{2-x}\le 2\Rightarrow \frac{3-2+x}{\sqrt{2-x}}\le 2\Rightarrow x+1\le 2\sqrt{2-x}$

Case $1:$ when $x+1\ge 0$
$\Rightarrow (x+1{)}^2\le 4(2-x)\Rightarrow x^2+6x-7\le 0\Rightarrow (x-1\left)\right(x+7)\le 0$
$\Rightarrow x\in [-7,1]\therefore x\in [-1,1]\cdots \left(2\right)$

Case $2:$ when $x+1<0$
$\Rightarrow x+1<0\Rightarrow x<-1\cdots \left(3\right)$

Hence from $\left(1\right),\left(2\right),\left(3\right)$ solution set will be
$(-\infty ,1]$