Difference between revisions of "1967 AHSME Problems/Problem 18"
(Created page with "== Problem == If <math>x^2-5x+6<0</math> and <math>P=x^2+5x+6</math> then <math>\textbf{(A)}\ P \; \text{can take any real value} \qquad \textbf{(B)}\ 20<P<30\\ \textbf{(C)}\ 0<...") |
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== Solution == | == Solution == | ||
| − | <math>\fbox{B}</math> | + | We are given that <math>x^2 - 5x + 6 < 0</math>, which, when factored, gives <math>(x - 2)(x-3) < 0</math>. This has a solution of <math>2<x<3</math>, because the original quadratic is <math>\cup</math>-shaped, and thus dips below the x-axis between the roots. |
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| + | Since <math>x^2 + 5x + 6</math> has a vertex minimum at <math>x = -\frac{5}{2}</math>, so it is increasing on the interval <math>[2, 3]</math>. Thus, evaluating <math>P</math> at <math>x=2</math> and <math>x=3</math> will give our bounds, and doing so gives <math>20 < P < 30</math>, or <math>\fbox{B}</math>. | ||
== See also == | == See also == | ||
Revision as of 02:09, 13 July 2019
Problem
If 
and 
then
Solution
We are given that 
, which, when factored, gives 
. This has a solution of 
, because the original quadratic is 
-shaped, and thus dips below the x-axis between the roots.
Since 
has a vertex minimum at 
, so it is increasing on the interval ![$[2, 3]$](http://latex.artofproblemsolving.com/7/5/f/75f22ce6d33ad4e7448cfea33d2bb3e6d944d883.png)
. Thus, evaluating 
at 
and 
will give our bounds, and doing so gives 
, or 
.
See also
| 1967 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
