Question

The roots of a quadratic equation a x +b x +c =0 are (2+i √2)/3 and (2−i √2)/3 . Find the values of b and c if a = −1.

Answer 1

`\begin{cases} x=(2+i√(2))/(3)\implies 3x=2+i√(2)\implies 3x-2-i√(2)=0\\\\ x=(2-i√(2))/(3)\implies 3x=2-i√(2)\implies 3x-2+i√(2)=0 \end{cases} \\\\\\ \stackrel{ \textit{original polynomial} }{a(3x-2-i√(2))(3x-2+i√(2))=\stackrel{ 0 }{y}} \\\\[-0.35em] ~\dotfill`

`\stackrel{ \textit{difference of squares} }{[(3x-2)-(i√(2))][(3x-2)+(i√(2))]}\implies (3x-2)^2-(i√(2))^2 \\\\\\ (9x^2-12x+4)-(2i^2)\implies 9x^2-12x+4-(2(-1)) \\\\\\ 9x^2-12x+4+2\implies 9x^2-12x+6 \\\\[-0.35em] ~\dotfill\\\\ a(9x^2-12x+6)=y\hspace{5em}\stackrel{\textit{now let's make}}{a=-(1)/(9)} \\\\\\ -\cfrac{1}{9}(9x^2-12x+6)=y\implies \boxed{-x^2+\cfrac{4}{3}x-\cfrac{2}{3}=y}`