1951 AHSME Problems/Problem 26

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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 \[(-1)^2-4(-m^2-m)=0\] \[4m^2+4m+1=0\] \[(2m+1)^2=0\] \[m=-\frac{1}{2} \Rightarrow \boxed{\textbf{(E)}}\]

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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