Question

If $\frac{x+\sqrt{x+2}}{x-\sqrt{x+2}}\ge 1,$ then

• A. $x\in [-2,-1]\cup [2,\infty )$
• B. $x\in [-2,-1)\cup [2,\infty )$
• C. $x\in [-2,\infty )-\{-2,1\}$
• D. $x\in (2,\infty )\cup \{-2\}$

Answer 1

The correct option is D $x\in (2,\infty )\cup \{-2\}$

$\frac{x+\sqrt{x+2}}{x-\sqrt{x+2}}\ge 1,$ to exists
$x+2\ge 0,x-\sqrt{x+2}\ne 0\Rightarrow x\ge -2,x^2\ne x+2\Rightarrow x\ge -2,x^2-x-2\ne 0\Rightarrow x\ge -2,(x-2)(x+1)\ne 0\Rightarrow x\in [-2,\infty )-\{-1,2\}\cdots \left(1\right)$

Now
$\frac{x+\sqrt{x+2}}{x-\sqrt{x+2}}-1\ge 0\Rightarrow \frac{2\sqrt{x+2}}{x-\sqrt{x+2}}\ge 0\Rightarrow \frac{1}{x-\sqrt{x+2}}>0,\text{or}x=-2[\because \sqrt{x+2}\ge 0]\Rightarrow x-\sqrt{x+2}>0,\text{or}x=-2\Rightarrow x>\sqrt{x+2},\text{or}x=-2$
$\because \sqrt{x+2}\ge 0\Rightarrow x>0$

Now $x^2>x+2,x>0,\text{or}x=-2$
$\Rightarrow x^2-x-2>0,x>0,\text{or}x=-2\Rightarrow (x-2)(x+1)>0,x>0,\text{or}x=-2$
$\Rightarrow x\in (2,\infty )\cup \{-2\}\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$
$x\in (2,\infty )\cup \{-2\}$