# 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 $\fbox{A}$.

## See also

 1967 AHSME (Problems • Answer Key • Resources) Preceded byProblem 16 Followed byProblem 18 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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