Question

Solve the following inequality:
$\frac{2}{x}+3⩽\sqrt{41-\frac{16}{x}}.$

Answer 1

$\frac{2}{x}+3\le \sqrt{41-\frac{16}{x}}$

$41-\frac{16}{x}\ge ({\frac{2}{x}+3)}^2$

$41-\frac{16}{x}\ge \frac{4}{x^2}+9+\frac{12}{x}$

$\frac{4}{x^2}+\frac{28}{x}-32\le 0$

$\frac{4+28x-32x^2}{x^2}\le 0$

$32x^2-28x-4\ge 0$

$8x^2-7x-1\ge 0$

$(8x+1)(x-1)\ge 0$

$x\epsilon (-\infty ,\frac{-1}{8})\bigcup (1,\infty )$