Problem:
Suppose that a and b are nonzero real numbers, and that the equation x2+ax+b=0 has solutions a and b. Then the pair (a,b) is
Answer Choices:
A. (β2,1)
B. (β1,2)
C. (1,β2)
D. (2,β1)
E. (4,4)
Solution:
The given conditions imply that
x2+ax+b=(xβa)(xβb)=x2β(a+b)x+ab,
so
a+b=βa and ab=b
Since bξ =0, the second equation implies that a=1. The first equation gives b=β2, so (a,b)=(1,β2).
Answer: Cβ.
The problems on this page are the property of the MAA's American Mathematics Competitions