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

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Solution

The correct option is B
2

$x_{1} + x_{2} = \frac{- b}{a}, x_{1} x_{2} = \frac{c}{a} \frac{1}{x_{1}^{2}} + \frac{1}{x_{2}^{2}} = \frac{x_{1}^{2} + x_{2}^{2}}{x_{1}^{2} x_{2}^{2}} = \frac{(x_{1} + x_{2})^{2} - 2 x_{1} x_{2}}{(x_{1} x_{2})^{2}} = {(\frac{x_{1} + x_{2}}{x_{1} x_{2}})}^{2} - \frac{2}{x_{1} x_{2}} = \frac{b^{2}}{c^{2}} - \frac{2 a}{c} N o w, \frac{b^{2}}{c^{2}} - \frac{2 a}{c} = \frac{- b}{a} . . . (G i v e n) \Rightarrow \frac{b^{2}}{c} - \frac{2 a c}{c} = \frac{- b c}{a} \Rightarrow \frac{b^{2}}{a c} - 2 = \frac{- b c}{a^{2}} \Rightarrow \frac{b^{2}}{a c} + \frac{b c}{a^{2}} = 2$

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If the sum of the roots of the equation ax2+bx+c=0 is equal to the sum of the squares of their reciprocals, then b2ac+bca2=

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