1963 AHSME Problems/Problem 14

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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 $\boxed{\textbf{(A)}}$.

See Also

1963 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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