Question

Solve the following inequality:
$\frac{\sqrt{x+20}}{x}-1<0.$

Answer 1

$\frac{\sqrt{x+20}}{x}-1<0$ $x+20>0$ ,$x\ge -20$ ,$x\in (-20,\infty )$

$\frac{\sqrt{x+20}}{x}<1$

squaring both sides

$\frac{x+20}{x^2}-1<0$

$\frac{x+20-x^2}{x^2}<0$

$\frac{x^2-x-20}{x^2}>0$

$\frac{(x-5)(x+4)}{x^2}>0$

$x\in (-\infty ,-4)\cup (5,\infty )$

$\therefore x\in (-20,4)\cup (5,\infty )$