If $\alpha ,\beta $ be the roots of the equation $2{x^2} - 2({m^2} + 1)x + ({m^4} + {m^2} + 1) = 0$ then $({\alpha ^2} + {\beta ^2}) = $
A) $1$
B) ${m^2}$
C) $0$
D) $m$

Hint: In this question, we have to find the roots of the equation for which we have to consider the sum of roots and after which we consider the product of roots by putting the value of alpha and beta we get the result. As by doing comparison of equation as $a{x^2} + bx + c = 0$ we can easily complete it.

Formula Used: Sum of roots $(\alpha + \beta ) = \dfrac{{ - b}}{a}$
Product of roots $(\alpha \beta ) = \dfrac{c}{a}$
$({\alpha ^2} + {\beta ^2}) = {(\alpha + \beta )^2} - 2\alpha \beta $

Complete step by step Solution:
As in the question, the given equation is,
$2{x^2} - 2({m^2} + 1)x + ({m^4} + {m^2} + 1) = 0$
By doing comparison as for $a{x^2} + bx + c = 0$ we get the respective values of a, b and c as,
$a = 2,b = - 2({m^2} + 1)$ and $c = ({m^4} + {m^2} + 1)$ ,
Now, for the Sum of roots, we get,
$(\alpha + \beta ) = \dfrac{{ - [ - 2({m^2} + 1)]}}{2}$
By doing further solutions we get the solution of the sum of roots as,
$(\alpha + \beta ) = ({m^2} + 1)......(i)$
Similarly, as for the product of roots, we get,
$(\alpha \beta ) = \dfrac{{{m^4} + {m^2} + 1}}{2}.....(ii)$
Now, according to question we have to find the $({\alpha ^2} + {\beta ^2})$
We know that,
$({\alpha ^2} + {\beta ^2}) = {(\alpha + \beta )^2} - 2\alpha \beta $
Now, by putting the values from the Equations we get the solution as follows,
$({\alpha ^2} + {\beta ^2}) = {({m^2} + 1)^2} - 2\left( {\dfrac{{{m^4} + {m^2} + 1}}{2}} \right)$
By doing further solutions we get,
$({\alpha ^2} + {\beta ^2}) = {m^4} + 2{m^2} + 1 - {m^4} - {m^2} - 1$
By completing the calculation, we get,
$({\alpha ^2} + {\beta ^2}) = {m^2}$
Therefore, the correct answer is ${m^2}$.

Therefore, the correct option is (B).

Note:The important point about the quadratic equation and roots are as follows: The intersection of an equation's graph with the x-axis at its roots, which is the physical significance of the roots. In the Cartesian plane, the real line is represented by the x-axis. As a result, the equation cannot be factorized if it has unreal roots because it won't cross the x-axis.