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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given the equation $x^4-x^2+x-1=0$.
Let $\alpha = \frac{1+\sqrt{3}i}{2}$ be a root.
Since the coefficients are real,the conjugate $\beta = \frac{1-\sqrt{3}i}{2}$ must also be a root.
Sum of roots $\alpha+\beta = \frac{1+\sqrt{3}i}{2} + \frac{1-\sqrt{3}i}{2} = 1$.
Product of roots $\alpha\beta = \left(\frac{1+\sqrt{3}i}{2}\right)\left(\frac{1-\sqrt{3}i}{2}\right) = \frac{1+3}{4} = 1$.
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:
$(x^2-x+1)(x^2+x-1) = 0$.
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 $x = \frac{-1 \pm \sqrt{1+4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.
Note: The original question statement implies the roots are $1, -1$,but the polynomial $x^4-x^2+x-1$ factors as $(x-1)(x+1)(x^2+1)$ is incorrect. Let us re-evaluate: $x^4-x^2+x-1 = x^2(x^2-1) + (x-1) = (x-1)(x^2(x+1)+1) = (x-1)(x^3+x^2+1)$.
Actually,$x^4-x^2+x-1 = (x^2-x+1)(x^2+x-1)$.
The real roots are $\frac{-1 \pm \sqrt{5}}{2}$.

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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