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(D) Given the quadratic equation $2x^2 - 2(m^2 + 1)x + m^4 + m^2 + 1 = 0$.Comparing with $ax^2 + bx + c = 0$,we have $a = 2$,$b = -2(m^2 + 1)$,and $c

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$m^2$

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(D) Given the quadratic equation $2x^2 - 2(m^2 + 1)x + m^4 + m^2 + 1 = 0$.Comparing with $ax^2 + bx + c = 0$,we have $a = 2$,$b = -2(m^2 + 1)$,and $c = m^4 + m^2 + 1$.Sum of roots: $\alpha + \beta = -\frac{b}{a} = \frac{2(m^2 + 1)}{2} = m^2 + 1$.Product of roots: $\alpha \beta = \frac{c}{a} = \frac{m^4 + m^2 + 1}{2}$.We know that $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$.Substituting the values:$\alpha^2 + \beta^2 = (m^2 + 1)^2 - 2 \left( \frac{m^4 + m^2 + 1}{2} \right)$.$\alpha^2 + \beta^2 = (m^4 + 2m^2 + 1) - (m^4 + m^2 + 1)$.$\alpha^2 + \beta^2 = m^4 - m^4 + 2m^2 - m^2 + 1 - 1 = m^2$.

If $\alpha$ and $\beta$ are the roots of the equation $2x^2 - 2(m^2 + 1)x + m^4 + m^2 + 1 = 0$,then find the value of $\alpha^2 + \beta^2$.

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