Difference between revisions of "MIE 2016/Problem 2"
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(e) <math>k\geq8</math> | (e) <math>k\geq8</math> | ||
| − | ==Solution 2== | + | ==Solution 1== |
| + | ===Objective:=== | ||
| + | We can solve this problem in two steps: First, we solve for the range of <math>\frac{x^2-2x-14}{x}>3</math>, then combine it with the range of <math>x\leq12</math> 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 <math>\frac{x^2-2x-14}{x}>3</math>. | ||
| + | We now simplify the inequality: | ||
| + | |||
| + | Case 1: <math>0<x\leq12</math> | ||
| + | |||
| + | <cmath>\frac{x^2-2x-14}{x}>3</cmath> | ||
| + | <cmath>x^2 - 2x - 14>3x</cmath> | ||
| + | <cmath>x^2 - 5x - 14 > 0</cmath> | ||
| + | Factoring, we get | ||
| + | <cmath>(x-7)(x+2)>0</cmath> | ||
| + | Now, <math>x</math> can be greater than <math>7</math> or less than <math>-2</math>. | ||
| + | But in this case, <math>0<x\leq12</math>, and this further restricts our solutions. | ||
| + | So, for the case where <math>0<x\leq12</math>, our solutions are <math>7<x\leq12</math> | ||
| + | |||
| + | Case 2: <math>x<0</math> | ||
| + | <cmath>\frac{x^2-2x-14}{x}>3</cmath> | ||
| + | <cmath>x^2-2x-14<3x</cmath> | ||
| + | <cmath>x^2-5x - 14<0</cmath> | ||
| + | <cmath>(x-7)(x+2)<0</cmath> | ||
| + | We have in this case that <math>-2<x<7</math>, but the case statement further restricts our solutions. | ||
| + | |||
| + | For this case, the solutions are <math>-2<x<0</math> | ||
| + | |||
| + | ===Step 2=== | ||
| + | Now, we know the solutions for <math>x</math>: in the first case, where <math>7<x\leq12</math>, the integer solutions are <math>x = {8, 9, 10, 11, 12}</math> | ||
| + | |||
| + | In the second case, where <math>-2<x<0</math>, the only integer solution is <math>x = {-1}</math> | ||
| + | |||
| + | The union of these two cases gives <math>x = {-1, 8, 9, 10, 11, 12}</math>. | ||
| + | |||
| + | There are <math>k=6</math> solutions and <math>6\leq k\leq8</math>, giving <math>\boxed{D)}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 11:07, 10 September 2020
Problem 2
The following system has 
integer solutions. We can say that:
(a) 
(b) 
(c) 
(d) 
(e) 
Solution 1
Objective:
We can solve this problem in two steps: First, we solve for the range of 
, then combine it with the range of 
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 .
We now simplify the inequality:
Case 1: 
![\[\frac{x^2-2x-14}{x}>3\]](http://latex.artofproblemsolving.com/d/9/2/d92e0685582338f36c13cd9709247ff8ed9fd37c.png)
![\[x^2 - 2x - 14>3x\]](http://latex.artofproblemsolving.com/7/8/d/78d2e5251ffb1fdbb69fbd0fd372427485d74e11.png)
![\[x^2 - 5x - 14 > 0\]](http://latex.artofproblemsolving.com/1/e/d/1ed72d5256d83391738c6a9f937ca2cd678feb47.png)
Factoring, we get
![\[(x-7)(x+2)>0\]](http://latex.artofproblemsolving.com/1/5/9/1597dac3d5bfc8012479bab8a93f079bd5a4ae5b.png)
Now, 
can be greater than 
or less than 
.
But in this case, , and this further restricts our solutions.
So, for the case where , our solutions are 
Case 2: 
![\[x^2-2x-14<3x\]](http://latex.artofproblemsolving.com/1/e/3/1e3566d0472f90a6f5e965e0d424463584f47e84.png)
![\[x^2-5x - 14<0\]](http://latex.artofproblemsolving.com/4/8/6/486702f14059361b599583e9f3855a7099de0a65.png)
![\[(x-7)(x+2)<0\]](http://latex.artofproblemsolving.com/d/a/a/daa7db1470c28d303e5d4c86a71f803338cb5f62.png)
We have in this case that 
, but the case statement further restricts our solutions.
For this case, the solutions are 
Step 2
Now, we know the solutions for : in the first case, where , the integer solutions are 
In the second case, where , the only integer solution is 
The union of these two cases gives 
.
There are 
solutions and , giving 
