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(B) Given the quadratic equation $x^2+bx+c=0$,where $\alpha$ and $\beta$ are the roots.From the relation between roots and coefficients,we have $\alph

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$-1$

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(B) Given the quadratic equation $x^2+bx+c=0$,where $\alpha$ and $\beta$ are the roots.From the relation between roots and coefficients,we have $\alpha+\beta = -b$ $(i)$ and $\alpha\beta = c$ (ii).We know that $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$.Substituting the given values: $5 = (-b)^2 - 2c$,which gives $2c = b^2-5$ or $c = \frac{b^2-5}{2}$.We also know that $\alpha^3+\beta^3 = (\alpha+\beta)(\alpha^2+\beta^2 - \alpha\beta)$.Substituting the given values: $9 = (-b)(5 - c)$.Substitute $c = \frac{b^2-5}{2}$ into the equation: $9 = -b(5 - \frac{b^2-5}{2})$.$9 = -b(\frac{10-b^2+5}{2}) = -b(\frac{15-b^2}{2})$.$18 = -15b + b^3$,which simplifies to $b^3 - 15b - 18 = 0$.Testing for roots,if $b=-3$: $(-3)^3 - 15(-3) - 18 = -27 + 45 - 18 = 0$. So,$b=-3$ is a root.Then $c = \frac{(-3)^2-5}{2} = \frac{9-5}{2} = 2$.Thus,$b+c = -3+2 = -1$.Therefore,option $(b)$ is correct.

If $\alpha, \beta$ are the roots of the quadratic equation $x^2+bx+c=0$ such that $\alpha^2+\beta^2=5$ and $\alpha^3+\beta^3=9$,then $b+c=$

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