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(D) Given the equation $x^2 - 4\sqrt{2}kx + 2e^{4\ln k} - 1 = 0$.Since $e^{4\ln k} = k^4$,the equation becomes $x^2 - 4\sqrt{2}kx + 2k^4 - 1 = 0$.From

Vedclass · Accepted Answer

$-280$

Vedclass · Answer

(D) Given the equation $x^2 - 4\sqrt{2}kx + 2e^{4\ln k} - 1 = 0$.Since $e^{4\ln k} = k^4$,the equation becomes $x^2 - 4\sqrt{2}kx + 2k^4 - 1 = 0$.From the properties of roots,$\alpha + \beta = 4\sqrt{2}k$ and $\alpha\beta = 2k^4 - 1$.We are given $\alpha^2 + \beta^2 = 66$.Using the identity $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$,we have:$(4\sqrt{2}k)^2 = 66 + 2(2k^4 - 1)$$32k^2 = 66 + 4k^4 - 2$$4k^4 - 32k^2 + 64 = 0$Dividing by $4$,we get $k^4 - 8k^2 + 16 = 0$,which is $(k^2 - 4)^2 = 0$.Thus,$k^2 = 4$,so $k = 2$ or $k = -2$.Now,$\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta) = (4\sqrt{2}k)(66 - (2k^4 - 1)) = (4\sqrt{2}k)(67 - 2k^4)$.If $k = 2$,$\alpha^3 + \beta^3 = (4\sqrt{2}(2))(67 - 2(16)) = (8\sqrt{2})(67 - 32) = (8\sqrt{2})(35) = 280\sqrt{2}$.If $k = -2$,$\alpha^3 + \beta^3 = (4\sqrt{2}(-2))(67 - 2(16)) = (-8\sqrt{2})(35) = -280\sqrt{2}$.Since the question implies a single value and $-280\sqrt{2}$ is an option,we select $D$.

If $\alpha$ and $\beta$ are roots of the equation $x^2 - 4\sqrt{2}kx + 2e^{4\ln k} - 1 = 0$ for some $k$,and $\alpha^2 + \beta^2 = 66$,then $\alpha^3 + \beta^3$ is equal to: (in $\sqrt{2}$)

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