1952 AHSME Problems/Problem 23

Revision as of 22:53, 24 January 2014 by Throwaway1489 (talk | contribs)

Problem

If $\frac{x^2-bx}{ax-c}=\frac{m-1}{m+1}$ has roots which are numerically equal but of opposite signs, the value of $m$ must be:

$\textbf{(A) \ }\frac{a-b}{a+b} \qquad \textbf{(B) \ }\frac{a+b}{a-b} \qquad \textbf{(C) \ }c \qquad \textbf{(D) \ }\frac{1}{c} \qquad \textbf{(E) \ }1$

Solution

Cross-multiplying, we find that $(m+1)x^2-(bm+am+b-a)x+c(m-1)=0$. Because the roots of this quadratic are $r_1$ and $-r_1$, their sum is $0$. According to Vieta's Formulas, $\frac{bm+am+b-a}{m+1}=0$, or $m(a+b)=a-b$. Hence, $m=\boxed{\textbf{(A)}\ \frac{a-b}{a+b}}$.

See also

1952 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png