If the sum of the roots of the quadratic equa | vedclass

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(A) Let $\alpha$ and $\beta$ be the roots of the quadratic equation $ax^2 + bx + c = 0$.From the properties of roots,we have $\alpha + \beta = -\frac{

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$2$

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(A) Let $\alpha$ and $\beta$ be the roots of the quadratic equation $ax^2 + bx + c = 0$.From the properties of roots,we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.According to the problem,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}$$\alpha + \beta = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2}$Substitute $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$:$-\frac{b}{a} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}$$-\frac{b}{a} = \frac{(-\frac{b}{a})^2 - 2(\frac{c}{a})}{(\frac{c}{a})^2}$$-\frac{b}{a} = \frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\frac{c^2}{a^2}}$Multiply both sides by $\frac{c^2}{a^2}$:$-\frac{b}{a} \cdot \frac{c^2}{a^2} = \frac{b^2}{a^2} - \frac{2c}{a}$$-\frac{bc^2}{a^3} = \frac{b^2}{a^2} - \frac{2c}{a}$Divide the entire equation by $\frac{c}{a}$ (assuming $c 
eq 0$):$-\frac{bc}{a^2} = \frac{b^2}{ac} - 2$Rearranging the terms,we get:$\frac{b^2}{ac} + \frac{bc}{a^2} = 2$.

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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