Difference between revisions of "1954 AHSME Problems/Problem 25"
Katzrockso (talk | contribs) (Created page with "== Problem 25== The two roots of the equation <math>a(b-c)x^2+b(c-a)x+c(a-b)=0</math> are <math>1</math> and: <math> \textbf{(A)}\ \frac{b(c-a)}{a(b-c)}\qquad\textbf{(B)}\ ...") |
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== Solution == | == Solution == | ||
Let the other root be <math>k</math>. Then by Vieta's Formulas, <math>k\cdot 1=\frac{c(a-b)}{a(b-c)}</math>, <math>\fbox{D}</math> | Let the other root be <math>k</math>. Then by Vieta's Formulas, <math>k\cdot 1=\frac{c(a-b)}{a(b-c)}</math>, <math>\fbox{D}</math> | ||
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| + | ==See Also== | ||
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| + | {{AHSME 50p box|year=1954|num-b=24|num-a=26}} | ||
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| + | {{MAA Notice}} | ||
Latest revision as of 00:29, 28 February 2020
Problem 25
The two roots of the equation 
are 
and:
Solution
Let the other root be 
. Then by Vieta's Formulas, 
, 
See Also
| 1954 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Problem 26 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
