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 $r_1$ 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, $m=\boxed{\textbf{(A)}\ \frac{a-b}{a+b}}$ .

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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1952 AHSME Problems/Problem 23

Problem

Solution

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