How to solve the inequality $(-K^2+13K+44)/(14-K)>0$?

Question

The inequality to solve: $$\left[\frac{-K^2+13K+44}{14-K}\right] > 0$$

How do I solve this? I tried this: $$ -K^2+13K+44 > 0 \quad \text{(multiply both sides by $14-K$)}\\ K^2-13K < 44\\ K(K-13) < 44 $$ Is this correct? Any way to get a more precise $K$ value? Thanks.

Answer 1

Hints You have 3 cases: $14-K < 0, 14-K >0$, and the easy one $14-K = 0$. In the first two, you end up with a different sign after multiplication. Take the one you used ($14-K>0$).

Then indeed $-K^2+13K+44>0$ but if you factor the left-hand side to get $(K-a)(K-b)<0$ for some $a,b$ you can find. When is a product negative? Can you take it from here?

Answer 2

Hint:

In order to maintain the same direction of the inequality you can multiply both sides by the positive term $(14-K)^2$.

Thus you will obtain the following inequality

$(14-K)(-K^2+13K+44)>0$

If you solve the quadratic $-K^2+13K+44=0$, you will obtain two roots $a$ and $b$ - I'll leave this for you to work it out.

Once you have found $a$ and $b$, you will be able to factorise $-K^2+13K+44=(a+K)(b-K)$, resulting in the inequality

$(14-K)(a+K)(b-K)>0$

Now, in order to satisfy the above inequality, either all three terms need to be positive, or two of the terms need to be negative and one term positive.

The task is then to find which values of $K$ will satisfy these conditions.