Question

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

Answer 1

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

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

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

$\frac{1-3x}{x^2}\ge 0$

$x<0or0<x\le \frac{1}{3}$

$\frac{1}{x}-\frac{1}{2}>0$

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

$0<x<2$

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

$\frac{1-3x}{x^2}<{(\frac{1}{x}-\frac{1}{2})}^2$

$\frac{1}{x^2}-\frac{3}{x}<\frac{1}{x^2}-\frac{1}{x}+\frac{1}{4}$

$-\frac{2}{x}-\frac{1}{4}<0$

$\frac{-8-x}{4x}<0$

$\frac{x+8}{4x}>0$

$x<-8orx>0$

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$0<x\le \frac{1}{3}$

$\sqrt{\frac{1}{x^2}-\frac{3}{x}}<\frac{1}{x}-\frac{1}{2}:\left[\begin{array}{cc}Solution:& 0<x\le \frac{1}{3}\\ Decimal:& 0<x\le 0.33333\dots \\ IntervalNotation:& (0,\frac{1}{3}]\end{array}\right]$