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        riddles >> putnam exam (pure math) >> Re: An algebra II problem. Easy for me and you but
        
(Message started by: towr on Dec 11^th, 2006, 12:33pm)

Title: Re: An algebra II problem. Easy for me and you but
Post by towr on Dec 11^th, 2006, 12:33pm

I seem to get imaginary numbers as well. Any reason why you oughtn't get them? (Like it's not in the curriculum or something?)

And you're certain about all the signs and stuff in the original problem?

Title: Re: An algebra II problem. Easy for me and you but
Post by Sameer on Dec 11^th, 2006, 12:44pm

Isn't this the same problem you posted earlier with quadratic roots of equation Ax^2+Bx+C=0 being A and B, C being the discriminant?

Title: Re: An algebra II problem. Easy for me and you but
Post by towr on Dec 11^th, 2006, 1:33pm

Well, there's two options.
1) Check the equations again. However it is also possible the teacher made a mistake, it happens.
2) Do the work as best as possible even though the result doesn't make sense. The important thing is that you understand the method, and apply it correctly. If the problem is faulty, you can still apply the method correctly.
You might go as far as to suggest where a typo might have led the problem to go astray. (I think I did that on a test once, and got extra points for spotting, fixing and solving the problem)

Title: Re: An algebra II problem. Easy for me and you but
Post by towr on Dec 11^th, 2006, 2:06pm

For the record, I get A = [-5 +/- sqrt(-7) ]/8
(And B = (1 + 4A) A^2, C = (1 + 4A) A^4 )
And my algebra program gets the same.

If we had (1-4A) instead of where we have (1+4A) you would get a real values.
Or alternatively if we have A + B = B / A, then we get rational values. Which test-technically is even better, imo.
There's more options though, A * B = - C / A would also yield rational values.