# Math

What is the sum of all values of k such that the equation 2x^2-kx+8=0 has two distinct integer solutions?

Thank you for your help! :)

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1. two distinct solutions means that the discriminant is positive. So, we need

k^2 - 64 > 0

also, we need

(k±√(k^2-64))/4 to be an integer.

So, k±√(k^2-64) must be a multiple of 4. and k^2-64 must be an integer.

so, what pythagorean triples do you know that have

a^2+8^2 = k^2

6,8,10: 10±6 = 4,16
15,8,17: 17±15 = 2,32

I guess 10 is the only value. 17 has one non-integer solution.

Maybe k can be negative.

Maybe there are some fractional values, but I can't think of any.

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posted by Steve

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