If k>0 and the product of the roots of equation x to power 2-3kx+2e power 2logk-1=0 is 7 then sum of roots is?

Answer 1

Hmmm. I hope you mean

x^2 - 3kx +2e^(2logk-1) = 0

as always, for ax^2+bx+c, the sum of the roots is -b/a.
Here, that would be 3k

Now, 2e^(2logk-1) = 2* 1/e * e^(2logk) = 2/e k^2

So, the product of the roots is c/a = 2/e k^2 = 7
That means k=√(7e/2)

Thus, the sum of the roots is 3√(7e/2)