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: \begin{align*} \frac{h}{2} = r+s = 4 &\rightarrow h = 8 \\ \frac{2k}{2} = rs = -3 &\rightarrow k = -3 \\ \end{align*}

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

1957 AHSC (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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.

Page

Toolbox

Search

1957 AHSME Problems/Problem 2

Problem

Solution

See also