1957 AHSME Problems/Problem 2

Problem

In the equation $2x^2 - hx + 2k = 0$, the sum of the roots is $4$ and the product of the roots is $-3$. Then $h$ and $k$ have the values, respectively:

$\textbf{(A)}\ 8\text{ and }{-6} \qquad  \textbf{(B)}\ 4\text{ and }{-3}\qquad  \textbf{(C)}\ {-3}\text{ and }4\qquad \textbf{(D)}\ {-3}\text{ and }8\qquad \textbf{(E)}\ 8\text{ and }{-3}$

Solution

Let the roots of the given equation be $r$ and $s$. Then, by Vieta's Formulas, we have the following: h2=r+s=4h=82k2=rs=3k=3

Thus, our answer is $\boxed{\textbf{(E) } 8 \text{ and } -3}$.

See also

1957 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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