Question

If the roots of the equation $(m-2)x^2-(8-2m)x-(8-3m)=0$ are real and opposite in sign, then the number of integral value(s) of $m$ is

Answer 1

$(m-2)x^2-(8-2m)x-(8-3m)=0$
$m-2\ne 0$ as the equation has two roots.
$\Rightarrow x^2-\left(\frac{8-2m}{m-2}\right)x-\left(\frac{8-3m}{m-2}\right)=0$
Let $f\left(x\right)=x^2-\left(\frac{8-2m}{m-2}\right)x-\left(\frac{8-3m}{m-2}\right)$
Since, roots are opposite in sign, hence $0$ lies between the roots.
$\Rightarrow f\left(0\right)<0$
$\Rightarrow -\left(\frac{8-3m}{m-2}\right)<0$
$\Rightarrow (m-2)(3m-8)<0$
$\Rightarrow m\in (2,\frac{8}{3})$
Hence, no integral value of $m$ in this interval.