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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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Verified Answer
The correct answer is:
$1,-1$
We have,
Let
$$
\begin{gathered}
x^4-x^2+x-1=0 \\
\alpha=\frac{1+\sqrt{3} i}{2}, \beta=\frac{1-\sqrt{3} i}{2} \\
\alpha+\beta=\frac{1+\sqrt{3} i}{2}+\frac{1-\sqrt{3} i}{2}=1 \\
\alpha \beta=\left(\frac{1+\sqrt{3} i}{2}\right)\left(\frac{1-\sqrt{3} i}{2}\right)=\frac{1+3}{4}=1
\end{gathered}
$$
Then equation is $x^2-(\alpha+\beta) x+\alpha \beta=0$
$$
\Rightarrow \quad x^2-x+1=0
$$
Now, divide $x^4-x^2+x-1$ by $x^2-x+1$,
We get, $x^2-1$
$\therefore$ The real roots are $x^2-1=0$
$$
x= \pm 1
$$
Let
$$
\begin{gathered}
x^4-x^2+x-1=0 \\
\alpha=\frac{1+\sqrt{3} i}{2}, \beta=\frac{1-\sqrt{3} i}{2} \\
\alpha+\beta=\frac{1+\sqrt{3} i}{2}+\frac{1-\sqrt{3} i}{2}=1 \\
\alpha \beta=\left(\frac{1+\sqrt{3} i}{2}\right)\left(\frac{1-\sqrt{3} i}{2}\right)=\frac{1+3}{4}=1
\end{gathered}
$$
Then equation is $x^2-(\alpha+\beta) x+\alpha \beta=0$
$$
\Rightarrow \quad x^2-x+1=0
$$
Now, divide $x^4-x^2+x-1$ by $x^2-x+1$,
We get, $x^2-1$
$\therefore$ The real roots are $x^2-1=0$
$$
x= \pm 1
$$
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