Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 12"
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== Problem == | == Problem == | ||
+ | If the equations <math> (1) x^2 + ax + b = 0</math> and <math> (2) x^2 + cx + d = 0 </math> have exactly one root in common, and <math> abcd\ne 0,</math> then the other root of equation <math> (2) </math> is | ||
− | <center><math> \mathrm{(A) \ } \qquad \mathrm{(B) \ } \qquad \mathrm{(C) \ } \qquad \mathrm{(D) \ } \qquad \mathrm{(E) \ } </math></center> | + | <center><math> \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 </math></center> |
== Solution == | == Solution == | ||
+ | Let <math>(1)</math> have roots <math>x=1,2</math> and <math>(2)</math> have roots <math>x=1,3</math>. Thus: | ||
+ | <math>(1)</math> <math>x^{2}-3x+2=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 choice must equal <math>3</math>. The answer is <math>(a)</math>. |
− | * [[University of South Carolina High School Math Contest/1993 Exam]] | + | |
+ | ---- | ||
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+ | * [[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]] | ||
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+ | [[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
Solution
Let have roots and have roots . Thus:
Thus, we know that and our answer choice must equal . The answer is .