Question

Find the value of $x$ for which following is true -

$sgn(x^2-2x-8)=-1$

Answer 1

$sgn\left(x\right)=\frac{\left|x\right|}{x},$ when $x\ne 0.$

&

$sgn\left(x\right)=0$ when $x=0.$

$sgn(x^2-2x-8)=$ $\frac{|x^2-2x-8|}{x^2-2x-8},$ when $x\ne 0.$

$=0,$ when $x=0.$

$=\frac{|x^2-4x+2x-8|}{x^2-4x+2x-8}$

$=0$

$=\frac{\left|\right(x-4\left)\right(x+2\left)\right|}{\left|\right(x-4\left)\right(x+2\left)\right|}$ when $x\ne 0$

$=0$ when $x=0$

Hence it is clear that when $x\in (2,4)$

$sgn(x^2-2x-8)=-1$