Difference between revisions of "1951 AHSME Problems/Problem 26"
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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 == | ||
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
![\[xm(x-1)-m(m+1)=x(x-1)(m-1)\]](http://latex.artofproblemsolving.com/e/0/b/e0bcd122541b34f8e8ac078283d2e688c6f742d1.png)
![\[x^2m-xm-m^2-m=x^2m-xm-x^2+x\]](http://latex.artofproblemsolving.com/0/6/d/06d685cb88a5d9bc74501f179a05546ffb1df13b.png)
![\[-m^2-m=-x^2+x\]](http://latex.artofproblemsolving.com/d/a/f/dafbebac279bdeba7d903968e5a92e762261894f.png)
![\[x^2-x-(m^2+m)=0\]](http://latex.artofproblemsolving.com/e/3/1/e31f52223fd65876efbe93b99e0c215e9be9dcc0.png)
The roots of this quadratic are equal if and only if its discriminant
(
) evaluates to 0.
This means
![\[(-1)^2-4(-m^2-m)=0\]](http://latex.artofproblemsolving.com/e/3/2/e32bc768ded7d64eb4a2abc6e09530bebb9afb4c.png)
![\[4m^2+4m+1=0\]](http://latex.artofproblemsolving.com/8/5/6/8564107c00478b28db878c2f35cba36a7143e2c3.png)
![\[(2m+1)^2=0\]](http://latex.artofproblemsolving.com/7/e/e/7ee95f992bf5e816d88d91228b47c384dee014df.png)
![\[m=-\frac{1}{2} \Rightarrow \boxed{\textbf{(E)}}\]](http://latex.artofproblemsolving.com/4/8/1/481379f1f6b299dcf8f2bb3a9d6563d843eb17f7.png)
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 25 |
Followed by Problem 27 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
