set of values of $k$ for which $kx^2-(4-2k)x-8<0$ for exactly $3$ integers values of $x$

Question

The set of values of $k$ for which $kx^2-(4-2k)x-8<0$ for exactly $3$ integers values of $x$ is

Try: $$kx^2-4x+2kx-8<0\Rightarrow x(kx-4)+2(kx-4)<0$$

So $$(x+2)(kx-4)<0$$

Could some help me to solve it, Thanks

Answer 1

Obviously $k>0$, otherwise there will be too many negative integers. Then from your factorisation, the inequality is satisfied iff $x\in (-2, 4/k)$, so $k\in [2,4)$ gives you exactly three integers for $x$.