1951 AHSME Problems/Problem 26

Problem

In the equation $\frac {x(x - 1) - (m + 1)}{(x - 1)(m - 1)} = \frac {x}{m}$ the roots are equal when

$\textbf{(A)}\ m = 1\qquad\textbf{(B)}\ m =\frac{1}{2}\qquad\textbf{(C)}\ m = 0\qquad\textbf{(D)}\ m =-1\qquad\textbf{(E)}\ m =-\frac{1}{2}$

Solution

Multiplying both sides by $(x-1)(m-1)m$ gives us $xm(x-1)-m(m+1)=x(x-1)(m-1)$ $x^2m-xm-m^2-m=x^2m-xm-x^2+x$ $-m^2-m=-x^2+x$ $x^2-x-(m^2+m)=0$ The roots of this quadratic are equal if and only if its discriminant ( $b^2-4ac$ ) evaluates to 0. This means $(-1)^2-4(-m^2-m)=0$ $4m^2+4m+1=0$ $(2m+1)^2=0$ $m=-\frac{1}{2} \Rightarrow \boxed{\textbf{(E)}}$

1951 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 25	Followed by Problem 27
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1951 AHSME Problems/Problem 26

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