Question

Let $\alpha$, $\alpha^2$ be the roots of $x^2 + x + 1 = 0$, then the equation whose roots are $\alpha^{31}$, $\alpha^{62}$ is

• A. $x^2-x+1=0$
• B. $x^2+x-1=0$
• C. $x^2+x+1=0$
• D. $x^{60}+x^{30}+1=0$

Answer 1

Answer: (C)

Given equation is $x^{2}+x+1=0$. Since, $\alpha, \alpha^{2}$ are the roots of the equation.
$\therefore \alpha+\alpha^{2}=-1 \,\,\,\ldots(i)$
and $\alpha^{3}=1 \,\,\,\ldots(i i)$
Now, for the equation of roots are $\alpha^{31}$ and $\alpha^{62}$.
$\alpha^{31}+\alpha^{62}=\alpha^{31}\left(1+\alpha^{31}\right)$
$\Rightarrow \alpha^{31}+\alpha^{62}=\alpha^{30} \alpha\left(1+\alpha^{30} \cdot \alpha\right)$
$\Rightarrow \alpha^{31}+\alpha^{62}=\left(\alpha^{3}\right)^{10} \cdot\left\{1+\alpha^{30} \cdot \alpha\right\}$
$\Rightarrow \alpha^{31}+\alpha^{62}=\left(\alpha^{3}\right)^{10} \cdot \alpha\left\{1+\left(\alpha^{3}\right)^{10} \cdot \alpha\right\}$
$\Rightarrow \alpha^{31}+\alpha^{62}=\alpha(1+\alpha) \,\,\,\,$ [using Eq. (ii)]
$\Rightarrow \alpha^{31}+\alpha^{62}=-1 \,\,\,\,$ [using Eq. $\left.(i)\right]$
Again $\alpha^{31} \cdot \alpha^{62}=\alpha^{93}$
$=\left(\alpha^{3}\right)^{31}=1$
$\therefore $ Required equation is
$x^{2}-\left(\alpha^{31}+\alpha^{62}\right) x+\alpha^{31} \cdot \alpha^{62}=0$
$\Rightarrow x^{2}+x+1=0$