Question

Solve the following inequalities
$\sqrt{25-x^2}⩽\frac{12}{x}.$

Answer 1

$\sqrt{25-x^2}\le \frac{12}{x}$

Square both sides

$\Rightarrow 25-x^2\le \frac{144}{x^2}$

$\Rightarrow x^4-25x^2+144\ge 0$

Let $u=x^2$ and $u^2=x^4$

$\Rightarrow u^2-25u+144\ge 0$

$\Rightarrow (u-16)(u-9)\ge 0$

$\Rightarrow (x^2-16)(x^2-9)\ge 0$

$\Rightarrow (x-4)(x+4)(x-3)(x+3)\ge 0$

For $x<-4$

$(x-4)(x+4)(x-3)(x+3)>0$

at $x=-4$, it will be zero.

For $-4<x<-3$,

$(x-4)(x+4)(x-3)(x+3)<0$

For $-3\le x\le 3$

$(x-4)(x+4)(x-3)(x+3)\ge 0$

Similarly

for $x\ge 4$

$(x-4)(x+4)(x-3)(x+3)\ge 0$

Solution : - $(-\infty ,-4]\cup [-3,3]\cup [4,\infty )$.