2x^2 −6x+k = 0 where k<0 ((α/β) + (β/α))


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Question Number 192928 by BaliramKumar last updated on 31/May/23

$2 x^{2} - 6 x + k = 0 w h e r e k < 0$ ${(\frac{α}{β} + \frac{β}{α})}_{\max} = ?$
	Answered by MM42 last updated on 31/May/23

	$s = 3 & p = \frac{k}{2}$ $A = \frac{α^{2} + β^{2}}{α β} = \frac{s^{2} - 2 p}{p} = \frac{9 - k}{\frac{k}{2}} = - 2 + \frac{18}{k} \overset{k < 0}{\to} A < - 2$ $\Rightarrow i f k < 0 \Rightarrow s u p A = - 2 b u t m a x A d o s e n o t e x i s t$
	Commented by BaliramKumar last updated on 31/May/23

	$thanks Sir$
	Commented by MM42 last updated on 31/May/23

	$g o o d l u c k$
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