Question

Let $a=\cos \frac{2\pi }{7}+i\sin \frac{2\pi }{7},\alpha =a+a^2+a^4$ and $\beta =a^3+a^5+a^6$. Then, the equation whose roots are $\alpha ,\beta$ 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

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

We have, $a=\cos \frac{2\pi }{7}+i\sin \frac{2\pi }{7}$
Therefore, $a^7={[\cos \frac{2\pi }{7}+i\sin \frac{2\pi }{7}]}^7=\cos 2\pi +i\sin 2\pi =1+0i=1$
Now, $\alpha +\beta =a+a^2+a^3+a^4+a^5+a^6$
$=a\left\{\frac{1-a^6}{1-a}\right\}=\frac{a-a^7}{1-a}=\frac{a-1}{1-a}[\because a^7=1]$
and $\alpha \beta =(a+a^2+a^4)(a^3+a^5+a^6)$
$=a^4+a^5+a^6+3a^7+a^8+a^9+a^{10}=3+a+a^2+a^3+a^4+a^5+a^6[\because a^7=1,a^8=a^7.a=a;a^9=a^7.a^2=a^2;a^{10}=a^7.a^3=a^3]$
$=3+\frac{a-1}{1-a}=3-1=2$
So, the required equation is
$x^2-(\alpha +\beta )x+\alpha \beta =0\Rightarrow x^2+x+2=0$