Difference between revisions of "1969 AHSME Problems/Problem 14"
m (→See also) |
Rockmanex3 (talk | contribs) (Solution to Problem 14) |
||
| Line 10: | Line 10: | ||
== Solution == | == Solution == | ||
| − | <math>\ | + | Factor the [[difference of squares]]. |
| + | <cmath>\frac{(x+2)(x-2)}{(x+1)(x-1)}>0</cmath> | ||
| + | Note that the graph intersects the x-axis at when <math>x = \pm2</math> or <math>x \pm 1</math>, so check the sign of the result to see if it is positive. After testing, <math>x<-2</math> or <math>-1<x<1</math> or <math>x>2</math>, so the answer is <math>\boxed{\textbf{(A)}}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 04:35, 7 June 2018
Problem
The complete set of 
-values satisfying the inequality 
is the set of all such that:
Solution
Factor the difference of squares.
![\[\frac{(x+2)(x-2)}{(x+1)(x-1)}>0\]](http://latex.artofproblemsolving.com/e/5/4/e54da0d47f940c02f6ccc1cf9f2ec6d8f5fc3e41.png)
Note that the graph intersects the x-axis at when 
or 
, so check the sign of the result to see if it is positive. After testing, 
or 
or 
, so the answer is 
.
See also
| 1969 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
