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(B) Given that $\alpha$ and $\beta$ are roots of the quadratic equation $x^2-4x+5=0$. From the relation between roots and coefficients,we have $\alpha

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$x^2-10x+34=0$

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(B) Given that $\alpha$ and $\beta$ are roots of the quadratic equation $x^2-4x+5=0$. From the relation between roots and coefficients,we have $\alpha+\beta=4$ and $\alpha\beta=5$. Let the new roots be $S_1 = \alpha^2+\beta$ and $S_2 = \alpha+\beta^2$. The sum of the new roots is: $S_1+S_2 = (\alpha^2+\beta)+(\alpha+\beta^2) = (\alpha^2+\beta^2)+(\alpha+\beta)$ $= ((\alpha+\beta)^2-2\alpha\beta)+(\alpha+\beta)$ $= (4^2-2(5))+4 = (16-10)+4 = 6+4 = 10$. The product of the new roots is: $S_1 	imes S_2 = (\alpha^2+\beta)(\alpha+\beta^2) = \alpha^3+\alpha^2\beta^2+\alpha\beta+\beta^3$ $= (\alpha^3+\beta^3)+(\alpha\beta)^2+\alpha\beta$ $= ((\alpha+\beta)^3-3\alpha\beta(\alpha+\beta))+(\alpha\beta)^2+\alpha\beta$ $= (4^3-3(5)(4))+(5)^2+5$ $= (64-60)+25+5 = 4+30 = 34$. The required quadratic equation is $x^2-(S_1+S_2)x+(S_1 	imes S_2) = 0$. Substituting the values,we get $x^2-10x+34=0$.

If $\alpha$ and $\beta$ are the roots of the equation $x^2-4x+5=0$,then the quadratic equation whose roots are $\alpha^2+\beta$ and $\alpha+\beta^2$ is

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