# Difference between revisions of "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 $\fbox{B}$.