Question

Solve the following inequalities.
$\frac{1-\sqrt{1-8x^2}}{2x}<1$

Answer 1

$\Rightarrow \frac{1-\sqrt{1-8x^2}}{2x}<1$

$\Rightarrow 1-\sqrt{1-8x^2}<2x$

$\Rightarrow 1-2x<\sqrt{1-8x^2}$

Square both sides,

$\Rightarrow 1+4x^2-4x<1-8x^2$

$\Rightarrow 12x^2-4x<0$

$\Rightarrow 3x^2-x<0$

$\Rightarrow x(3x-1)<0$

at $x<0,x(3x-1)<0$

at $x=0,x(3x-1)=0$

at $0<x<\frac{1}{3},x(3x-1)<0$

at $x\ge \frac{1}{3},x(3x-1)\ge 0$

$\therefore solution:0<x<\frac{1}{3}$

Interval notation : $(0,\frac{1}{3})$.