Two complex numbers $a$ and $\beta$ are such

$\text{ that }\alpha +\beta =2\text{ and }{\alpha }^4+{\beta }^4=272\text{, then the quadratic equation }$

$\text{ whose roots are }\alpha \text{ and }\beta \text{ can be }$

          $\alpha +\beta =2\Rightarrow {\alpha }^2+{\beta }^2+2\alpha \beta =4$
$\begin{array}{l}\Rightarrow {\alpha }^2+{\beta }^2=4-2\alpha \beta \Rightarrow {\left({\alpha }^2+{\beta }^2\right)}^2=16-16\alpha \beta +4{\alpha }^2{\beta }^2\\ \Rightarrow {\alpha }^4+{\beta }^4+2{\alpha }^2{\beta }^2=16-16\alpha \beta +4{\alpha }^2{\beta }^2\\ \Rightarrow 272+2(\alpha \beta {)}^2=16-16\alpha \beta +4(\alpha \beta {)}^2\\ \Rightarrow 2(\alpha \beta {)}^2-16(\alpha \beta )-256=0\\ \Rightarrow (\alpha \beta {)}^2-8(\alpha \beta )-128=0\\ \Rightarrow (\alpha \beta -16)(\alpha \beta +8)=0\\ \Rightarrow \alpha \beta =16\text{ or }\alpha \beta =-8\end{array}$

Thus, required equation is either

or       $\begin{array}{l}{x}^2-2x+16=0\\ x^2-2x-8=0\end{array}$

$\therefore$ Required answer is (c).