Question

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

• A. $n^2$
• B. $-n^2$
• C. $n^4$
• D. $-n^4$

Answer 1

The correct option is A

$n^2$

Find the value of ${\mathit{\alpha }}^{\mathbf{2}}\mathbf{+}{\mathit{\beta }}^{\mathbf{2}}\mathbf{}$:

Given that $\alpha$and $\beta$are the roots of the equations$x^2-(1+n^2)x+\frac{1}{2}(1+n^2+n^4)=0$,

$\Rightarrow (\alpha +\beta )=1+n^2;\left(\alpha \beta \right)=\frac{(1+n^2+n^4)}{2}$

Now,

$$\begin{gathered} \Rightarrow ({\alpha }^2+{\beta }^2)={(1+n^2)}^2–2\times \frac{1}{2}(1+n^2+n^4) \\ \Rightarrow ({\alpha }^2+{\beta }^2)=1+n^4+2n^2–1–n^2–n^4 \\ \Rightarrow ({\alpha }^2+{\beta }^2)=n^2 \end{gathered}$$ $\mathbf{(}{\mathit{\alpha }}^{\mathbf{2}}\mathbf{+}\mathbf{}{\mathit{\beta }}^{\mathbf{2}}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{(}\mathit{\alpha }\mathbf{}\mathbf{+}\mathbf{}\mathit{\beta }\mathbf{)}}^{\mathbf{2}}\mathbf{–}\mathbf{}\mathbf{2}\mathit{\alpha }\mathit{\beta }$

Hence, the correct option is (A).