Asked by Swathi
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?
Answers
Answered by
Steve
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)
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)
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