If alpha_1, alpha_2 are the roots of x^2+ax+1 | vedclass

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(D) Given that $\alpha_1, \alpha_2$ are roots of $x^2+ax+1=0$,we have $\alpha_1+\alpha_2 = -a$ and $\alpha_1\alpha_2 = 1$.Given that $\alpha_3, \alpha

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$(b+a)^2$

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(D) Given that $\alpha_1, \alpha_2$ are roots of $x^2+ax+1=0$,we have $\alpha_1+\alpha_2 = -a$ and $\alpha_1\alpha_2 = 1$.Given that $\alpha_3, \alpha_4$ are roots of $x^2+bx+1=0$,we have $\alpha_3+\alpha_4 = -b$ and $\alpha_3\alpha_4 = 1$.We need to evaluate $E = (\alpha_1+\alpha_3)(\alpha_2+\alpha_3)(\alpha_1+\alpha_4)(\alpha_2+\alpha_4)$.Rearranging the terms,we get $E = [(\alpha_1+\alpha_3)(\alpha_2+\alpha_3)] 	imes [(\alpha_1+\alpha_4)(\alpha_2+\alpha_4)]$.Expanding these,$E = (\alpha_1\alpha_2 + \alpha_3(\alpha_1+\alpha_2) + \alpha_3^2) 	imes (\alpha_1\alpha_2 + \alpha_4(\alpha_1+\alpha_2) + \alpha_4^2)$.Substituting the known values,$E = (1 - a\alpha_3 + \alpha_3^2) 	imes (1 - a\alpha_4 + \alpha_4^2)$.Since $\alpha_3$ is a root of $x^2+bx+1=0$,$\alpha_3^2+b\alpha_3+1=0$,so $\alpha_3^2+1 = -b\alpha_3$.Similarly,$\alpha_4^2+1 = -b\alpha_4$.Thus,$E = (-b\alpha_3 - a\alpha_3) 	imes (-b\alpha_4 - a\alpha_4) = (-(a+b)\alpha_3) 	imes (-(a+b)\alpha_4)$.Therefore,$E = (a+b)^2 \alpha_3\alpha_4 = (a+b)^2(1) = (a+b)^2$.

If $\alpha_1, \alpha_2$ are the roots of $x^2+ax+1=0$ and $\alpha_3, \alpha_4$ are the roots of $x^2+bx+1=0$,then $(\alpha_1+\alpha_3)(\alpha_2+\alpha_3)(\alpha_1+\alpha_4)(\alpha_2+\alpha_4) = $

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