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== | ||
− | + | Multiplying both sides by <math>(x-1)(m-1)m</math> gives us | |
+ | <cmath>xm(x-1)-m(m+1)=x(x-1)(m-1)</cmath> | ||
+ | <cmath>x^2m-xm-m^2-m=x^2m-xm-x^2+x</cmath> | ||
+ | <cmath>-m^2-m=-x^2+x</cmath> | ||
+ | <cmath>x^2-x-(m^2+m)=0</cmath> | ||
+ | The roots of this quadratic are equal if and only if its discriminant | ||
+ | (<math>b^2-4ac</math>) evaluates to 0. | ||
+ | This means | ||
+ | <cmath>(-1)^2-4(-m^2-m)=0</cmath> | ||
+ | <cmath>4m^2+4m+1=0</cmath> | ||
+ | <cmath>(2m+1)^2=0</cmath> | ||
+ | <cmath>m=-\frac{1}{2} \Rightarrow \boxed{\textbf{(E)}}</cmath> | ||
− | ==See Also== | + | == See Also == |
+ | {{AHSME 50p box|year=1951|num-b=25|num-a=27}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 10:49, 16 April 2014
Problem
In the equation the roots are equal when
Solution
Multiplying both sides by gives us The roots of this quadratic are equal if and only if its discriminant () evaluates to 0. This means
See Also
1951 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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