Question

lf $a=\cos \frac{2\pi }{7}+i\sin \frac{2\pi }{7},\alpha =a+a^2+a^4$ and $\beta =a^3+a^5+a^6$, then $\alpha ,\beta$ are the roots of the equation

• A. $x^2+x+1=0$
• B. $x^2+x+2=0$
• C. $x^2+2x+2=0$
• D. $x^2+2x+3=0$

Answer 1

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

$a=e^{i2\pi /7}{a}^7=1\alpha +\beta$

$=a+a^2+a^3+a^4+a^5+a^6=\frac{a(a^6-1)}{(a-1)}=\frac{(a^7-a)}{(a-1)}=\frac{(1-a)}{(a-1)}$
$=-1$
$\alpha \beta =(a+a^2+a^4)(a^3+a^5+a^6)=(a^4+a^6+a^7+a^5+a^7+a^8+a^7+a^9+a^{10})=(a^4+a^6+1+a^5+1+a+1+a^2+a^3)=(3+a+a^2+a^3+a^4+a^5+a^6)=(3-1)$

$=2$
The equation can be written as
$x^2-(\alpha +\beta )x+\alpha \beta =x^2+x+2$
Hence, option B is correct.