Difference between revisions of "1967 AHSME Problems/Problem 18"

Latest revision as of 00:40, 16 August 2023

Problem

If $x^2-5x+6<0$ and $P=x^2+5x+6$ then

$\textbf{(A)}\ P \; \text{can take any real value} \qquad \textbf{(B)}\ 20<P<30\\ \textbf{(C)}\ 0<P<20 \qquad \textbf{(D)}\ P<0 \qquad \textbf{(E)}\ P>30$

Solution

We are given that $x^2 - 5x + 6 < 0$ , which, when factored, gives $(x - 2)(x-3) < 0$ . This has a solution of $2<x<3$ , because the original quadratic is $\cup$ -shaped, and thus dips below the x-axis between the roots.

Since $x^2 + 5x + 6$ has a vertex minimum at $x = -\frac{5}{2}$ , so it is increasing on the interval $[2, 3]$ . Thus, evaluating $P$ at $x=2$ and $x=3$ will give our bounds, and doing so gives $20 < P < 30$ , or $\fbox{B}$ .

1967 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 14: / Line 14: @@
 == See also ==
-{{AHSME box|year=1967|num-b=17|num-a=19}}
+{{AHSME 40p box|year=1967|num-b=17|num-a=19}}
 [[Category:Introductory Algebra Problems]]
 {{MAA Notice}}

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Difference between revisions of "1967 AHSME Problems/Problem 18"

Latest revision as of 00:40, 16 August 2023

Problem

Solution

See also