Question

Solve the following inequality:
$\sqrt{4-x^2}+\frac{|x|}{x}⩾0.$

Answer 1

$4-x^2\ge 0$

$x^2-4\le 0$

$(x-2)(x+2)\le 0$

$x\in (-2,2)$

If $x>0$

$\frac{\left|x\right|}{x}=1$

If $x<0$

$\frac{\left|x\right|}{x}=-1$

If $x>0$

$\sqrt{4-x^2}+1\ge 0$

$\sqrt{4-x^2}\ge -1$

$4-x^2\ge 1$

$x^2-3\le 0$

$x\in (-\sqrt{3},\sqrt{3})$

$\therefore x\in (0,\sqrt{3})$

If $x<0$

$\sqrt{4-x^2}-1\ge 0$

$\sqrt{4-x^2}\ge 1$

Squaring

$4-x^2\ge 1$ $\therefore x\in [-\sqrt{3},\sqrt{3}]-0$

$x^2-3\le 0$

$x\in (-\sqrt{3},\sqrt{3})$

$\therefore x\in (-\sqrt{3},0)$