Question

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

The correct option is C

$x^2+x+1=0$

Explanation:

Step 1: Find the sum and product of the roots of the given equation.

It is given that, $\alpha$and${\alpha }^2$, are the roots of$x^2+x+1=0$

$$\begin{gathered} \therefore \alpha +{\alpha }^2=\frac{-\left(1\right)}{1}\{sumofroots\} \\ \Rightarrow \alpha (1+\alpha )=-1 \\ \alpha \times {\alpha }^2=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$1$}\right.=1\{Productofzeros\} \\ \Rightarrow {\alpha }^3=1 \end{gathered}$$

Step 2: Form a quadratic equation whose roots are given.

$$\begin{gathered} {\alpha }^{31}and{\alpha }^{62}aretherootsofrequiredequation. \\ \therefore {\alpha }^{31}+{\alpha }^{62}=sumofroots \\ \Rightarrow {\alpha }^{30}(\alpha +{\alpha }^2)={\left({\alpha }^3\right)}^{10}(\alpha +{\alpha }^2) \\ \Rightarrow {\left(1\right)}^{10}(-1)=-1...\left(1\right) \\ andalso, \\ {\alpha }^{31}.{\alpha }^{62}={\alpha }^{93}={\left({\alpha }^3\right)}^{31}=1...\left(2\right) \\ \therefore equationis \\ {x}^2-(-1)x+1=x^2+x+1 \end{gathered}$$

Hence, Option$\left(C\right)$ is the correct option.