Question

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

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

Answer 1

Answer: (D)

$x^2+x+2=0$

Given $a=\cos \left(\frac{2\pi }{7}\right)+i\sin \left(\frac{2\pi }{7}\right)$
$\Rightarrow a^7={[\cos \left(\frac{2\pi }{7}\right)+i\sin \left(\frac{2\pi }{7}\right)]}^7$
$\Rightarrow a^7=\cos 2\pi +i\sin 2\pi$
$a^7=\mathrm{1....}\left(i\right)$
Let $S=\alpha +\beta =(a+a^2+a^4)+(a^3+a^5+a^6)=a+a^2+a^3+a^4+a^5+a^6$
Since it is GP
$\therefore S=\frac{a(1-a^6)}{1-a}$
$\Rightarrow S=\frac{a-a^7}{1-a}$ (From (i))
$\Rightarrow S=\frac{a-1}{1-a}$
$\Rightarrow S=-\frac{(1-a)}{(1-a)}=-\mathrm{1...}\left(ii\right)$
Again
$P=\alpha \beta =(a+a^2+a^4)(a^3+a^5+a^6)=3+(a+a^2+a^3+a^4+a^5+a^6)$
$=3+S=3-1$ (From Eq.(ii))
$=\mathrm{2....}\left(iii\right)$
$\therefore$ Required equation is
$\therefore x^2-(\alpha +\beta )x+\alpha \beta =0$
$\therefore x^2+x+2=0$ (From Eqs. (ii) and (iii))