If alpha and beta are the roots of x^2-2x+4=0 | vedclass

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(C) Given,$\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$. From the properties of roots,we have: $\alpha+\beta = 2$ ...$(i)$ $\alpha\beta = 4$ ...$

Vedclass · Accepted Answer

$128$

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(C) Given,$\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$. From the properties of roots,we have: $\alpha+\beta = 2$ ...$(i)$ $\alpha\beta = 4$ ...$(ii)$ Now,we can write $\alpha^6+\beta^6 = (\alpha^2)^3 + (\beta^2)^3$. Alternatively,note that the roots of $x^2-2x+4=0$ are $x = \frac{2 \pm \sqrt{4-16}}{2} = 1 \pm \sqrt{3}i$. In polar form,$1 \pm \sqrt{3}i = 2(\frac{1}{2} \pm i\frac{\sqrt{3}}{2}) = 2e^{\pm i\pi/3}$. Thus,$\alpha = 2e^{i\pi/3}$ and $\beta = 2e^{-i\pi/3}$. Then,$\alpha^6 = (2e^{i\pi/3})^6 = 2^6 e^{i2\pi} = 64(1) = 64$. Similarly,$\beta^6 = (2e^{-i\pi/3})^6 = 2^6 e^{-i2\pi} = 64(1) = 64$. Therefore,$\alpha^6+\beta^6 = 64+64 = 128$.

If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$,then the value of $\alpha^6+\beta^6$ is

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