# 1967 AHSME Problems/Problem 18

## Problem

If $x^2-5x+6<0$ and $P=x^2+5x+6$ then $\textbf{(A)}\ P \; \text{can take any real value} \qquad \textbf{(B)}\ 2030$

## 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, 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 $\fbox{B}$.

## See also

 1967 AHSME (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 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

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