Question

let $a=e^{i\frac{2\pi }{13}}$ then the quadratic equation whose roots are $\alpha =a+a^3+a^4+a^{-4}+a^{-3}+a^{-1},\beta =a^2+a^5+a^6+a^{-6}+b^{-5}+a^{-2}$ is given by

• A. $x^2-x-3=0$
• B. $x^2-x+2=0$
• C. $x^2+x+3=0$
• D. $x^2+x-3=0$

Answer 1

The correct option is D $x^2+x-3=0$

Given $a=e^{i\frac{2\pi }{3}}$

Given $\alpha =a+a^3+a^4+a^{-4}+a^{-3}+a^{-1}$ and $\beta =a^2+a^5+a^6+a^{-6}+a^{-5}+a^{-2}$

We get $\alpha +\beta =a^{-6}+a^{-5}+a^{-4}+a^{-3}+a^{-2}+a^{-1}+a^1+a^2+a^3+a^4+a^5+a^6=-1$

$\Rightarrow \alpha \beta =3(a^{-6}+a^{-5}+a^{-4}+a^{-3}+a^{-2}+a^{-1}+a^1+a^2+a^3+a^4+a^5+a^6)=-3$

The quadratic equation whose roots are $\alpha ,\beta$ are, $x^2-(\alpha +\beta )x+\alpha \beta =0$

$\Rightarrow x^2+x-3=0$