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 $\fbox{A}$ .

1967 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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1967 AHSME Problems/Problem 17

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