Difference between revisions of "1954 AHSME Problems/Problem 25"

(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 01:29, 28 February 2020

Problem 25

The two roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are $1$ and:

$\textbf{(A)}\ \frac{b(c-a)}{a(b-c)}\qquad\textbf{(B)}\ \frac{a(b-c)}{c(a-b)}\qquad\textbf{(C)}\ \frac{a(b-c)}{b(c-a)}\qquad\textbf{(D)}\ \frac{c(a-b)}{a(b-c)}\qquad\textbf{(E)}\ \frac{c(a-b)}{b(c-a)}$

Solution

Let the other root be $k$. Then by Vieta's Formulas, $k\cdot 1=\frac{c(a-b)}{a(b-c)}$, $\fbox{D}$

See Also

1954 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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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