# Difference between revisions of "MIE 2016/Problem 2"

## Problem 2

The following system has $k$ integer solutions. We can say that:

$\begin{cases}\frac{x^2-2x-14}{x}>3\\\\x\leq12\end{cases}$

(a) $0\leq k\leq 2$

(b) $2\leq k\leq 4$

(c) $4\leq k\leq6$

(d) $6\leq k\leq8$

(e) $k\geq8$

## Solution 1

### Objective:

We can solve this problem in two steps: First, we solve for the range of $\frac{x^2-2x-14}{x}>3$, then combine it with the range of $x\leq12$ to get a compound inequality which we can use to find all possible integer solutions.

### Step 1

We first find the range of the inequality $\frac{x^2-2x-14}{x}>3$.

We now simplify the inequality:

Case 0: $x=0$

This has no solutions since $x=0$ will make the function undefined.

Case 1: $0

$$\frac{x^2-2x-14}{x}>3$$ $$x^2 - 2x - 14>3x$$ $$x^2 - 5x - 14 > 0$$ Factoring, we get $$(x-7)(x+2)>0$$ Now, $x$ can be greater than $7$ or less than $-2$. But in this case, $0, and this further restricts our solutions. So, for the case where $0, our solutions are $7

Case 2: $x<0$ $$\frac{x^2-2x-14}{x}>3$$ $$x^2-2x-14<3x$$ $$x^2-5x - 14<0$$ $$(x-7)(x+2)<0$$ We have in this case that $-2, but the case statement further restricts our solutions.

For this case, the solutions are $-2

### Step 2

Now, we know the solutions for $x$: in the first case, where $7, the integer solutions are $x = {8, 9, 10, 11, 12}$

In the second case, where $-2, the only integer solution is $x = {-1}$

The union of these two cases gives $x = {-1, 8, 9, 10, 11, 12}$.

There are $k=6$ solutions and $6\leq k\leq8$, giving $\boxed{D)}$ ~Windigo

## Solution 2

Because $x \le 12$ is the simpler condition, we can apply it to the solution of $\frac{x^2-2x-14}{x} < 3$. We can find the solution of the first inequality given in the problem by simplifying and using the Snake Method. To use the Snake Method, we need to have $0$ on one of the sides, and factor the other. To do this, we can subtract $\frac{3x}{x}$ from both sides, going from $$\frac{x^2-2x-14}{x} < 3$$ to $$\frac{x^2-2x-14-3x}{x} < 0$$ and combining like terms to get $$\frac{x^2-5x-14}{x} < 0$$ factoring to get the usable form of $$\frac{(x+2)(x-7)}{x} < 0$$ Using the snake method, we can build a table.

 $\space$ $x<-2$ $-2 $0 $7 $x+2$ $-$ $+$ $+$ $+$ $x$ $-$ $-$ $+$ $+$ $x-7$ $-$ $-$ $-$ $+$ multiplied $-$ $+$ $-$ $+$

Using this table, we can find out that this inequality is true when $-2 or $7. Applying the second inequality, we can modify these to $-2 or $7 so they fit the whole set. Counting the number of integers that satisfy these conditions, we can see that the six numbers $-1, 8, 9, 10, 11, 12$ fit these conditions. Therefore, the answer is $\boxed{D)}$. -Aurora64 ¯\_(ツ)_/¯