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?
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)
To find the sum of the roots of the quadratic equation, you can use the fact that for a quadratic equation of the form ax^2 + bx + c = 0, the sum of the roots is equal to -b/a.
In this case, the quadratic equation is x^2 - 3kx + 2e^(2logk-1) = 0.
We are given that the product of the roots is 7. Let's denote the roots as alpha and beta, so alpha * beta = 7.
Using the quadratic formula, the sum of the roots can be found as:
sum of roots = alpha + beta = (-(-3k))/1 = 3k.
Therefore, the sum of the roots is 3k.