First slide

Theory of equations

Question

$If the roots of the equation a x^{2} - b x + c = 0 are α, β then the roots of the equation b^{2} c x^{2} - a b^{2} x + a^{3} = 0 are$

Moderate

A

$\frac{1}{α^{3} + α β}, \frac{1}{β^{3} + α β}$
B

$\frac{1}{α^{2} + α β}, \frac{1}{β^{2} + α β}$
C

$\frac{1}{α^{4} + α β}, \frac{1}{β^{4} + α β}$
D

None of these

Solution

$Multiplying the second equation by \frac{c}{a^{3}}, we get \frac{b^{2} c^{2}}{a^{3}} x^{2} - \frac{b^{2} c}{a^{2}} x + c = 0$
$\begin{array}{l} \Rightarrow a {(\frac{b c}{a^{2}} x)}^{2} - b (\frac{b c}{a^{2}}) x + c = 0 w h i c h i s i n t h e f o r m o f t h e g i v e n e q u a t i o n w h o s e r o o t s a r e α, β \\ \Rightarrow \frac{b c}{a^{2}} x = α, β \\ \frac{b}{a} \cdot \frac{c}{a} x = α, β \\ f r o m t h e g i v e n e q u a t i o n α + β = \frac{b}{a}, α β = \frac{c}{a} \\ \Rightarrow (α + β) α β x = α, β \\ (α + β) α β x = α, (α + β) α β x = β \\ \Rightarrow x = \frac{1}{(α + β) α}, \frac{1}{(α + β) β} \end{array}$

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