Question

Solve the following inequality:
$\sqrt{1-\frac{x+2}{x^2}}<\frac{2}{3}.$

Answer 1

$\sqrt{1-\frac{x+2}{x^2}}<\frac{2}{3}$

$\sqrt{\frac{x^2-(x+2)}{x^2}}<\frac{2}{3}$

$\frac{x^2-(x+2)}{x^2}\ge 0$

$\frac{(x+1)(x-2)}{x^2}\ge 0$

$x\le -1orx\ge 2$

$\sqrt{\frac{x^2-(x+2)}{x^2}}<\frac{2}{3}$

$\frac{x^2-x-2}{x^2}<\frac{4}{9}$

$1-\frac{1}{x}-\frac{2}{x^2}<\frac{4}{9}$

$-\frac{2}{x^2}-\frac{1}{x}+\frac{5}{9}<0$

$\frac{-18-9x+5x^2}{9x^2}<0$

$\frac{(x-3)(5x+6)}{9x^2}<0$

$-\frac{6}{5}<x<0or0<x<3$

$-\frac{6}{5}<x\le -1or2\le x<3$

$\sqrt{1-\frac{x+2}{x^2}}<\frac{2}{3}:\left[\begin{array}{cc}Solution:& -\frac{6}{5}<x\le -1or2\le x<3\\ Decimal:& -1.2<x\le -1or2\le x<3\\ IntervalNotation:& (-\frac{6}{5},-1]\cup [2,3)\end{array}\right]$