If frac{1+sqrt{3}i}{2} is a root of the equat | vedclass

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(D) Given equation is $x^4-x^2+x-1=0$.Let $\alpha = \frac{1+\sqrt{3}i}{2}$. Since the coefficients are real,the conjugate $\beta = \frac{1-\sqrt{3}i}{

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$\frac{-1 \pm \sqrt{5}}{2}$

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(D) Given equation is $x^4-x^2+x-1=0$.Let $\alpha = \frac{1+\sqrt{3}i}{2}$. Since the coefficients are real,the conjugate $\beta = \frac{1-\sqrt{3}i}{2}$ must also be a root.Sum of roots $\alpha + \beta = \frac{1+\sqrt{3}i + 1-\sqrt{3}i}{2} = 1$.Product of roots $\alpha \beta = \frac{1^2 - (\sqrt{3}i)^2}{4} = \frac{1+3}{4} = 1$.The quadratic factor corresponding to these roots is $x^2 - (\alpha+\beta)x + \alpha\beta = x^2 - x + 1 = 0$.Dividing $x^4-x^2+x-1$ by $x^2-x+1$,we get $x^4-x^2+x-1 = (x^2-x+1)(x^2+x-1) = 0$.The real roots are obtained from $x^2+x-1=0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $x = \frac{-1 \pm \sqrt{1-4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.

If $\frac{1+\sqrt{3}i}{2}$ is a root of the equation $x^4-x^2+x-1=0$,then its real roots are:

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