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.
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! :)
2 answers
the answer is zero yw