# 1952 AHSME Problems/Problem 23

## 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 additive inverses, 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 (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 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. Invalid username
Login to AoPS