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(C) Given the quadratic equation ${x^2} - (1 + {n^2})x + \frac{1}{2}(1 + {n^2} + {n^4}) = 0$.For a quadratic equation $ax^2 + bx + c = 0$,the sum of r

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${n^2}$

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(C) Given the quadratic equation ${x^2} - (1 + {n^2})x + \frac{1}{2}(1 + {n^2} + {n^4}) = 0$.For a quadratic equation $ax^2 + bx + c = 0$,the sum of roots $\alpha + \beta = -b/a$ and the product of roots $\alpha \beta = c/a$.Here,$\alpha + \beta = 1 + {n^2}$ and $\alpha \beta = \frac{1}{2}(1 + {n^2} + {n^4})$.We know that ${\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} - 2\alpha \beta$.Substituting the values:${\alpha ^2} + {\beta ^2} = {(1 + {n^2})^2} - 2 	imes \frac{1}{2}(1 + {n^2} + {n^4})$${\alpha ^2} + {\beta ^2} = (1 + {n^4} + 2{n^2}) - (1 + {n^2} + {n^4})$${\alpha ^2} + {\beta ^2} = 1 + {n^4} + 2{n^2} - 1 - {n^2} - {n^4}$${\alpha ^2} + {\beta ^2} = {n^2}$.

If $\alpha$ and $\beta$ are the roots of the equation ${x^2} - (1 + {n^2})x + \frac{1}{2}(1 + {n^2} + {n^4}) = 0$,then the value of ${\alpha ^2} + {\beta ^2}$ is:

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