Question

Solve the following inequality:
$\sqrt{x}-3⩽\frac{2}{\sqrt{x}}-2\cdot$

Answer 1

$\sqrt{x}-3\le \frac{2}{\sqrt{x}}-2$ $x\ge 0$

$\sqrt{x}-\frac{2}{\sqrt{x}}\le 1$

squaring bothsides

$x-4+\frac{4}{x}\le 1$

$\frac{x^2-5x+4}{x}\le 0$

$\frac{(x-4)(x-1)}{x}\le 0$

$x\in (-\infty ,0)\cup [1,4]$

$\therefore x\in [1,4]$