If product of roots of the equation ${x^2} - 3kx + 2{e^{\log k}} - 1 = 0$ is $7,$ then
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roots are integers and positive
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roots are integers and negative
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roots are rational not integers
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roots are irrational
The product of roots for ax^2 + bx + c = 0 is c/a. Here, c = 2*e^(log k) - 1 = 2k - 1. So (2k - 1)/1 = 7, which means 2k = 8, so k = 4. The equation becomes x^2 - 12x + 7 = 0. The discriminant is 144 - 28 = 116, which is not a perfect square, so the roots are irrational.
Using e^(log k) = k, the given equation becomes x^2 - 3kx + 2k - 1 = 0. The product of the roots is given as 2k - 1 = 7, which means 2k = 8 and k = 4. Substituting k = 4 into the original equation yields x^2 - 12x + 7 = 0. The discriminant of this equation is (-12)^2 - 4(1)(7) = 144 - 28 = 116. Since 116 is not a perfect square, the roots are irrational.