Question

If \(a=cos\left(\frac{2\pi}{7} \right )+i~sin\left(\frac{2\pi}{7} \right )\) then the quadratic equation whose roots are \(\alpha=a+a^2+a^4~and~\beta=a^3+a^5+a^6\) is

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$a=cos\left(\frac{2\pi }{7}\right)+isin\left(\frac{2\pi }{7}\right)a^7=[cos\left(\frac{2\pi }{7}\right)+isin\left(\frac{2\pi }{7}\right)]=cos2\pi +isin2\pi =1......\left(i\right)$
$S=\alpha +\beta =(a+a^2+a^4)+(a^3+a^5+a^6)S=a+a^2+a^3+a^4+a^5+a^6=\frac{a(1-a^6)}{1-6}S=\frac{a-a^7}{1-a}=\frac{a-1}{1-a}=-1...........\left(ii\right)P=\alpha \beta =(a+a^2+a^4)(a^3+a^5{+}^6)=a^4+a^6+a^7+a^5+a^7+a^8+a^7+a^9+a^{10}=a^4+a^6+1+a^5+1+a+1+a^2+a^3\left[Frome{q}^n\right(i\left)\right]=3+(a+a^2+a^3+a^4+a^5+a^6)=3+S=3-1=2\left[From\right(ii\left)\right]$
Required equation is, $x^2-Sx+P=0$
$\Rightarrow x^2+x+2=0$.