Difference between revisions of "1951 AHSME Problems/Problem 26"

(Created page with "==Problem== In the equation <math>\frac {x(x - 1) - (m + 1)}{(x - 1)(m - 1)} = \frac {x}{m}</math> the roots are equal when <math> \textbf{(A)}\ m = 1\qquad\textbf{(B)}\ m =\fr...")
 
 
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==Solution==
 
==Solution==
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Multiplying both sides by <math>(x-1)(m-1)m</math> gives us
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<cmath>xm(x-1)-m(m+1)=x(x-1)(m-1)</cmath>
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<cmath>x^2m-xm-m^2-m=x^2m-xm-x^2+x</cmath>
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<cmath>-m^2-m=-x^2+x</cmath>
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<cmath>x^2-x-(m^2+m)=0</cmath>
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The roots of this quadratic are equal if and only if its discriminant
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(<math>b^2-4ac</math>) evaluates to 0.
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This means
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<cmath>(-1)^2-4(-m^2-m)=0</cmath>
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<cmath>4m^2+4m+1=0</cmath>
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<cmath>(2m+1)^2=0</cmath>
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<cmath>m=-\frac{1}{2} \Rightarrow \boxed{\textbf{(E)}}</cmath>
  
==See Also==
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== See Also ==
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{{AHSME 50p box|year=1951|num-b=25|num-a=27}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 10:49, 16 April 2014

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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