The correct option is (A)\(x^2 + x + 2 = 0\)
Given, \(a = \cos\frac{2\pi}{7} + i \sin\frac{2\pi}{7}\)
\(\therefore a^7 = \cos 2\pi + i \sin 2\pi\)
\( [ \because e^{i\theta} = \cos \theta + i \sin \theta ]\) \(= 1\)
Also, \(\alpha = a + a^2 + a^4,\beta = a^3 + a^5 + a^6\) then the sum of roots, \(S = \alpha + \beta = a + a^2 + a^3 + a^4 + a^5 + a^6\)
\(\Rightarrow S = \frac{a(1 - a^6)}{1 - a} = \frac{a - a^7}{1 - a}\) \(= \frac{a - 1}{1 - a} = -1 [\because a^7 = 1]\) Product of the roots, \(P = \alpha \beta = (a + a^2 + a^4)(a^3 + a^5 + a^6)\)
\(= a^4 + a^5 + 1 + a^6 + 1 + a^2 + 1 + a + a^3 [\because a^7 = 1]\)
\(= 3 + (a + a^2 + a^3 + a^4 + a^5 + a^6) = 3 - 1 = 2\)
Hence, the required quadratic equation is
\(x^2 + x + 2 = 0\).