Difference between revisions of "1957 AHSME Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | |||
| + | Let the roots of the given equation be <math>r</math> and <math>s</math>. 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 <math>\boxed{\textbf{(E) } 8 \text{ and } -3}</math>. | ||
== See also == | == See also == | ||
| − | {{AHSME 50p box|year=1957| | + | {{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 
, the sum of the roots is 
and the product of the roots is 
.
Then 
and 
have the values, respectively:
Solution
Let the roots of the given equation be 
and 
. 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 
.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
