$\begin{array}{l} I f t h e s u m o f t h e r o o t s o f t h e q u a d r a t i c e q u a t i o n a x^{2} + b x + c = 0 i s e q u a l t o t h e s u m o f t h e s q u a r e s o f t h e i r r e c i p r o c a l s \\ t h e n \frac{a}{c}, \frac{b}{a} a n d \frac{c}{b} a r e i n \end{array}$

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n a x^{2} + b x + c = 0 \\ L e t α, β a r e r o o t s t h e n α + β = \frac{- b}{a}, α β = \frac{c}{a} \\ G i v e n α + β = \frac{1}{α^{2}} + \frac{1}{β^{2}} \\ \Rightarrow α + β = \frac{α^{2} + β^{2}}{α^{2} β^{2}} \\ \Rightarrow α + β = \frac{{(α + β)}^{2} - 2 (\frac{c}{a})}{{(α β)}^{2}} \\ \Rightarrow \frac{- b}{a} = \frac{\frac{b^{2}}{a^{2}} - 2 (\frac{c}{a})}{\frac{c^{2}}{a^{2}}} \\ \Rightarrow \frac{- b}{a} (\frac{c^{2}}{a^{2}}) = \frac{b^{2}}{a^{2}} - 2 (\frac{c}{a}) \\ b c^{2} = - a b^{2} + 2 a^{2} c \\ \frac{b c^{2}}{a b c} = \frac{- a b^{2}}{a b c} + \frac{2 a^{2} c}{a b c} \\ \Rightarrow \frac{c}{a} = \frac{- b}{c} + \frac{2 a}{b} \\ 2 (\frac{a}{b}) = \frac{c}{a} + \frac{b}{c} \\ ∴ \frac{c}{a}, \frac{a}{b}, \frac{b}{c} a r e i n A . P . \end{array}$