1963 AHSME Problems/Problem 14

Problem

Given the equations $x^2+kx+6=0$ and $x^2-kx+6=0$ . If, when the roots of the equation are suitably listed, each root of the second equation is $5$ more than the corresponding root of the first equation, then $k$ equals:

$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ -5 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ -7 \qquad \textbf{(E)}\ \text{none of these}$

Solution

Let the two roots of $x^2 + kx + 6$ be $r$ and $s$ . By Vieta's Formulas, $r+s=-k$ Each root of $x^2 - kx + 6$ is five more than each root of the original, so using Vieta's Formula again yields $r+5+s+5 = k$ $r+s+10=k$ Substitute $r+s$ to get $-k + 10 = k$ $10 = 2k$ $5 = k$ The answer is $\boxed{\textbf{(A)}}$ .

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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1963 AHSME Problems/Problem 14

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