If the sum of the squares of the reciprocals | vedclass

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Question

(B) Given the quadratic equation $3x^{2} + \lambda x - 1 = 0$ with roots $\alpha$ and $\beta$.From the relation between roots and coefficients,we have

Vedclass · Accepted Answer

$24$

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(B) Given the quadratic equation $3x^{2} + \lambda x - 1 = 0$ with roots $\alpha$ and $\beta$.From the relation between roots and coefficients,we have $\alpha + \beta = -\frac{\lambda}{3}$ and $\alpha\beta = -\frac{1}{3}$.The sum of the squares of the reciprocals is given by $\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} = 15$.This simplifies to $\frac{\alpha^{2} + \beta^{2}}{(\alpha\beta)^{2}} = 15$,which is $\frac{(\alpha + \beta)^{2} - 2\alpha\beta}{(\alpha\beta)^{2}} = 15$.Substituting the values: $\frac{(-\lambda/3)^{2} - 2(-1/3)}{(-1/3)^{2}} = 15$.$\frac{\lambda^{2}/9 + 2/3}{1/9} = 15 \implies \lambda^{2} + 6 = 15 \implies \lambda^{2} = 9$.Now,we need to calculate $6(\alpha^{3} + \beta^{3})^{2}$.Recall that $\alpha^{3} + \beta^{3} = (\alpha + \beta)((\alpha + \beta)^{2} - 3\alpha\beta)$.$\alpha^{3} + \beta^{3} = (-\frac{\lambda}{3})((-\frac{\lambda}{3})^{2} - 3(-\frac{1}{3})) = (-\frac{\lambda}{3})(\frac{\lambda^{2}}{9} + 1)$.Since $\lambda^{2} = 9$,$\alpha^{3} + \beta^{3} = (-\frac{\lambda}{3})(\frac{9}{9} + 1) = (-\frac{\lambda}{3})(2) = -\frac{2\lambda}{3}$.Then $6(\alpha^{3} + \beta^{3})^{2} = 6(-\frac{2\lambda}{3})^{2} = 6(\frac{4\lambda^{2}}{9}) = 6(\frac{4 	imes 9}{9}) = 6 	imes 4 = 24$.

If the sum of the squares of the reciprocals of the roots $\alpha$ and $\beta$ of the equation $3x^{2} + \lambda x - 1 = 0$ is $15$,then $6(\alpha^{3} + \beta^{3})^{2}$ is equal to

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