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

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(C) Given that $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2-2x+4=0$. From the relation between roots and coefficients,we have: $\

Vedclass · Accepted Answer

$128$

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(C) Given that $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2-2x+4=0$. From the relation between roots and coefficients,we have: $\alpha+\beta = 2$ $(i)$ $\alpha\beta = 4$ $(ii)$ We can write $x^2-2x+4=0$ as $x^2-2x+1+3=0$,which implies $(x-1)^2 = -3$. Thus,$x-1 = \pm \sqrt{3}i$,so $x = 1 \pm \sqrt{3}i$. Let $\alpha = 1+\sqrt{3}i = 2(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 2e^{i\pi/3}$ and $\beta = 1-\sqrt{3}i = 2(\frac{1}{2} - i\frac{\sqrt{3}}{2}) = 2e^{-i\pi/3}$. 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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