Problem:
How many pairs (a,b) of non-zero real numbers satisfy the equation
a1β+b1β=a+b1β?
Answer Choices:
A. none
B. 1
C. 2
D. one pair for each bξ =0
E. two pairs for each bξ =0
Solution:
The given equation is equivalent to each of these equations
aba+bβ(a+b)2a2+ab+b2β=a+b1β=ab=0.β
Since the last equation is satisfied by the pairs (a,b) such that a=2βbΒ±β3b2ββ, there are no pairs of real numbers satisfying the original equation.