Solve the inequality $(1008-99K-0.75K^2)/(42-0.5K)>0$

Question

How do you solve $$\frac{1008-99K-0.75K^2}{42-0.5K} > 0$$ for $K$?

I don't think you can just get rid of the denominator by multiplying to the other side right?

Answer 1

You can get rid of the denominator as long as you assume $K \neq 84$, since then the denominator is zero. But you will have to keep track of whether the denominator is negative or positive.

So, assuming $K < 84$, the denominator is positive. Multiplying by $42 - 0.5K$ on both sides gives us

$$1008 - 99K - 0.75K^2 > 0$$

Solve this inequality and use only the values that lie in the range $K < 84$.

Assuming $K > 84$, the denominator is negative. Multiplying by $42 - 0.5K$ on both sides reverses the sign, and gives us:

$$1008 - 99K - 0.75K^2 < 0$$

The solution to this inequality that also satisfy $K>84$ are, together with the previous solutions to $K$, the full solution to the original problem.

Answer 2

Yes, I can (give you a hint): Why may the fraction change its sign only for $K\in\{-66\pm10\sqrt{57},84\}$? So the sign must be constant on the four intervals determined by those three values of $K$.