Difference between revisions of "1957 AHSME Problems/Problem 2"

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== Problem ==
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In the equation <math>2x^2 - hx + 2k = 0</math>, the sum of the roots is <math>4</math> and the product of the roots is <math>-3</math>.
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Then <math>h</math> and <math>k</math> have the values, respectively:
  
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<math>\textbf{(A)}\ 8\text{ and }{-6} \qquad
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\textbf{(B)}\ 4\text{ and }{-3}\qquad
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\textbf{(C)}\ {-3}\text{ and }4\qquad
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\textbf{(D)}\ {-3}\text{ and }8\qquad
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\textbf{(E)}\ 8\text{ and }{-3}  </math> 
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== Solution ==
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<math>\fbox{E}</math>
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== See also ==
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{{AHSME 50p box|year=1957|before=First Problem|num-a=2}}
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{{MAA Notice}}
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[[Category:AHSME]][[Category:AHSME Problems]]
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[[Category:Introductory Algebra Problems]]

Revision as of 11:02, 24 July 2024

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

$\fbox{E}$


See also

1957 AHSC (ProblemsAnswer KeyResources)
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

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