Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 12"
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<math>(2)</math> <math>x^{2}-4x+3=0</math> | <math>(2)</math> <math>x^{2}-4x+3=0</math> | ||
− | Thus, we know that <math>(a,b,c,d)=(-3,2,-4,3)</math> and our answer | + | Thus, we know that <math>(a,b,c,d)=(-3,2,-4,3)</math> and our answer choice must equal <math>3</math>. The answer is <math>(a)</math>. |
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− | * [[University of South Carolina High School Math Contest/1993 Exam]] | + | |
+ | * [[University of South Carolina High School Math Contest/1993 Exam/Problem 11|Previous Problem]] | ||
+ | * [[University of South Carolina High School Math Contest/1993 Exam/Problem 13|Next Problem]] | ||
+ | * [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]] | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 13:14, 12 October 2007
Problem
If the equations and
have exactly one root in common, and
then the other root of equation
is
![$\mathrm{(A) \ }\frac{c-a}{b-d}d \qquad \mathrm{(B) \ }\frac{a+c}{b+d}d \qquad \mathrm{(C) \ }\frac{b+c}{a+d}c \qquad \mathrm{(D) \ }\frac{a-c}{b-d} \qquad \mathrm{(E) \ }\frac{a+c}{b-d}c$](http://latex.artofproblemsolving.com/3/0/e/30e4811f03eb46eb9bea68efe4334a588bd1a071.png)
Solution
Let have roots
and
have roots
. Thus:
Thus, we know that and our answer choice must equal
. The answer is
.