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(A) Let $\alpha$ and $\beta$ be the roots of the equation $ax^2 + bx + c = 0$. Then,$\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.Gi

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$2$

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(A) Let $\alpha$ and $\beta$ be the roots of the equation $ax^2 + bx + c = 0$. Then,$\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.Given that the sum of the roots is equal to the sum of the squares of their reciprocals:$\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2}$$-\frac{b}{a} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2}$$-\frac{b}{a} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}$Substituting the values of $\alpha + \beta$ and $\alpha\beta$:$-\frac{b}{a} = \frac{(-\frac{b}{a})^2 - 2(\frac{c}{a})}{(\frac{c}{a})^2}$$-\frac{b}{a} = \frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\frac{c^2}{a^2}}$$-\frac{b}{a} = \frac{b^2 - 2ac}{a^2} 	imes \frac{a^2}{c^2}$$-\frac{b}{a} = \frac{b^2 - 2ac}{c^2}$$-bc^2 = a(b^2 - 2ac)$$-bc^2 = ab^2 - 2a^2c$$2a^2c = ab^2 + bc^2$Dividing both sides by $a^2c$:$2 = \frac{ab^2}{a^2c} + \frac{bc^2}{a^2c}$$2 = \frac{b^2}{ac} + \frac{bc}{a^2}$.

If the sum of the roots of the quadratic equation $ax^2 + bx + c = 0$ is equal to the sum of the squares of their reciprocals,then $\frac{b^2}{ac} + \frac{bc}{a^2} = $

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