• # question_answer If $a=\cos (2\pi /7)+i\,\sin (2\pi /7),$ then the quadratic equation whose roots are $\alpha =a+{{a}^{2}}+{{a}^{4}}$ and $\beta ={{a}^{3}}+{{a}^{5}}+{{a}^{6}}$ is [RPET 2000] A) ${{x}^{2}}-x+2=0$ B) ${{x}^{2}}+x-2=0$ C) ${{x}^{2}}-x-2=0$ D) ${{x}^{2}}+x+2=0$

$a=\cos (2\pi /7)+i\sin (2\pi /7)$ ${{a}^{7}}={{[\cos (2\pi /7)+i\sin (2\pi /7)]}^{7}}$     $=\,\cos 2\pi +i\sin 2\pi =1$             .....(i) $S\,\,=\alpha +\beta =(a+{{a}^{2}}+{{a}^{4}})+({{a}^{3}}+{{a}^{5}}+{{a}^{6}})$ $S=a+{{a}^{2}}+{{a}^{3}}+{{a}^{4}}+{{a}^{5}}+{{a}^{6}}$$=\frac{a(1-{{a}^{6}})}{1-a}$ $S=\,\,\frac{a-{{a}^{7}}}{1-a}=\frac{a-1}{1-a}=-1$          .....(ii) $P=\,\,\alpha \beta =(a+{{a}^{2}}+{{a}^{4}})\,({{a}^{3}}+{{a}^{5}}+{{a}^{6}})$ $={{a}^{4}}+{{a}^{6}}+{{a}^{7}}+{{a}^{5}}+{{a}^{7}}+{{a}^{8}}+{{a}^{7}}+{{a}^{9}}+{{a}^{10}}$ $={{a}^{4}}+{{a}^{6}}+1+{{a}^{5}}+1+a+1+{{a}^{2}}+{{a}^{3}}$ (From eqn (i)] $=3+(a+{{a}^{2}}+{{a}^{3}}+{{a}^{4}}+{{a}^{5}}+{{a}^{6}})$$=3+S$ $=$$3-1=2$ [From (ii)] Required equation is, ${{x}^{2}}-Sx+P=0$ Þ ${{x}^{2}}+x+2=0$.