Albert and Bernard
A and B are real numbers.
If A B = 2 0 1 5 , how many possible values are there for A + B ?
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2 solutions
Nice approach. (I think that you meant 2015 rather than 2014.)
I took the AM-GM approach, (just focusing on the first quadrant), in that
f ( A ) = A + A 2 0 1 5 ≥ 2 2 0 1 5 .
As f ( A ) is a continuous real-valued function for A = 0 , lim A → 0 + f ( A ) = ∞ , and since f ′ ( A ) = 1 − A 2 2 0 1 5 ≤ 0 for 0 < A ≤ 2 0 1 5 , at which point f ( A ) = 2 2 0 1 5 , we know that f ( A ) can take on any real value greater than 2 2 0 1 5 with A > 0 . For A < 0 we would have f ( A ) taking on any real value less than − 2 2 0 1 5 .
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Oh dear, looks like I'm still stuck in 2014 :p
That it can take on any complex value is a stronger result, but not necessary to solve the question. The AM-GM approach has its own strengths too, being constructive in a way in that it shows the range of f ( A ) for the different signs of A .
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Yes, the two approaches taken together paint a complete picture of the solution space. :)
Using AM-GM inequality, we have: (A+B)/2 >= sqrt(AB) (A+B)/2 >= sqrt(AB) so A+B >= 2*sqrt(2015) This means that (A+B) has infinite possible values. But (A+B) can be a real number, and the set of all real number is uncountable, so (A+B) has uncountably infinite possible values.
Consider the polynomial x 2 − k x + 2 0 1 5 which has roots A and B .
The constant term - and hence the product - is always 2 0 1 5 , but k = A + B can be any real (or even complex) number.