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 == |
Latest 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 . 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 | |
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All AHSME Problems and Solutions |
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