If the product of roots of the equation {x^2} | vedclass

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(B) The given equation is ${x^2} - 3kx + 2{e^{2\log k}} - 1 = 0$.Since ${e^{2\log k}} = {e^{\log {k^2}}} = {k^2}$,the equation becomes ${x^2} - 3kx +

Vedclass · Accepted Answer

$k = 2$

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(B) The given equation is ${x^2} - 3kx + 2{e^{2\log k}} - 1 = 0$.Since ${e^{2\log k}} = {e^{\log {k^2}}} = {k^2}$,the equation becomes ${x^2} - 3kx + 2{k^2} - 1 = 0$.The product of the roots of a quadratic equation $ax^2 + bx + c = 0$ is $\frac{c}{a}$.Here,the product of roots is $2{k^2} - 1$.Given that the product of roots is $7$,we have $2{k^2} - 1 = 7$,which implies $2{k^2} = 8$,so ${k^2} = 4$,giving $k = \pm 2$.Since the term $\log k$ exists in the original equation,$k$ must be positive,so $k = 2$.For the roots to be real,the discriminant $D = b^2 - 4ac$ must be $\ge 0$.$D = (-3k)^2 - 4(1)(2{k^2} - 1) = 9{k^2} - 8{k^2} + 4 = {k^2} + 4$.Since ${k^2} + 4 > 0$ for all real $k$,the roots are always real for any real $k$.However,given the options and the condition $k=2$,the roots are real when $k=2$.

If the product of roots of the equation ${x^2} - 3kx + 2{e^{2\log k}} - 1 = 0$ is $7$,then its roots will be real when:

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