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

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== Solution ==
 
== Solution ==
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<math>\fbox{E}</math>
 
  
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Let the roots of the given equation be <math>r</math> and <math>s</math>. Then, by [[Vieta's Formulas]], we have the following:
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\begin{align*}
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\frac{h}{2} = r+s = 4 &\rightarrow h = 8 \\
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\frac{2k}{2} = rs = -3 &\rightarrow k = -3 \\
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\end{align*}
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Thus, our answer is <math>\boxed{\textbf{(E) } 8 \text{ and } -3}</math>.
  
 
== See also ==
 
== See also ==
  
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{{AHSME 50p box|year=1957|before=First Problem|num-a=2}}
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{{AHSME 50p box|year=1957|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 
[[Category:AHSME]][[Category:AHSME Problems]]
 
[[Category:AHSME]][[Category:AHSME Problems]]
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Latest revision as of 10:06, 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

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 $\boxed{\textbf{(E) } 8 \text{ and } -3}$.

See also

1957 AHSC (Problems • Answer Key • Resources)
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

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