# Difference between revisions of "1967 AHSME Problems/Problem 17"

## Problem

If $r_1$ and $r_2$ are the distinct real roots of $x^2+px+8=0$, then it must follow that:

$\textbf{(A)}\ |r_1+r_2|>4\sqrt{2}\qquad \textbf{(B)}\ |r_1|>3 \; \text{or} \; |r_2| >3 \\ \textbf{(C)}\ |r_1|>2 \; \text{and} \; |r_2|>2\qquad \textbf{(D)}\ r_1<0 \; \text{and} \; r_2<0\qquad \textbf{(E)}\ |r_1+r_2|<4\sqrt{2}$

## Solution

We are given that the roots are real, so the discriminant is positive, which means $p^2 - 4(8)(1) > 0$. This leads to $|p| > 4\sqrt{2}$. By Vieta, the sum of the roots is $-p$, so we have $|-(r_1 + r_2)| \ge 4\sqrt{2}$, or $|r_1 + r_2| > 4\sqrt{2}$, which is option $\fbox{A}$.