1957 AHSME Problems/Problem 2

Revision as of 10:06, 24 July 2024 by Thepowerful456 (talk | contribs) (typo fix)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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
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 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png