If α,β are the roots of ax2+bx+c=0 then αβ-βα2=

α, β are roots of ax2+bx+c=0 thenα+β=−ba, αβ= ca∴ (αβ−βα)2=(α2−β2αβ)2= (α+β)2(α−β)2(αβ)2= (α+β)2[(α+β)2−4αβ](αβ)2=b2(b2−4ac)a2c2

If α,β are the roots of ax2+bx+c=0 then αβ-βα2=

α, β are roots of ax2+bx+c=0 thenα+β=−ba, αβ= ca∴ (αβ−βα)2=(α2−β2αβ)2= (α+β)2(α−β)2(αβ)2= (α+β)2[(α+β)2−4αβ](αβ)2=b2(b2−4ac)a2c2

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If $α, β$ are the roots of ${ax}^{2} + bx + c = 0$ then ${(\frac{α}{β} - \frac{β}{α})}^{2} =$

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$\frac{b^{2} (b^{2} - 4 ac)}{{ca}^{3}}$

$\frac{b^{2} (b^{2} - 4 ac)}{c^{2} a^{2}}$

$\frac{b^{2} (b^{2} - 4 ac)}{a^{4}}$

$\frac{b^{2} (b^{2} - 4 ac)}{c^{4}}$

answer is A.

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Detailed Solution

$\begin{array}{l} α, β a r e r o o t s o f a x^{2} + b x + c = 0 t h e n \\ α + β = \frac{- b}{a}, α β = \frac{c}{a} \\ ∴ {(\frac{α}{β} - \frac{β}{α})}^{2} = {(\frac{α^{2} - β^{2}}{α β})}^{2} \\ = \frac{{(α + β)}^{2} {(α - β)}^{2}}{{(α β)}^{2}} \\ = \frac{{(α + β)}^{2} [{(α + β)}^{2} - 4 α β]}{{(α β)}^{2}} \\ = \frac{b^{2} (b^{2} - 4 a c)}{a^{2} c^{2}} \end{array}$