If α and β are the roots of equation (√(t/(1−t))) + (√((1−t)/t)) = ((13)/6) . Find 2α+3β .


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Question Number 159188 by cortano last updated on 14/Nov/21

$I f α a n d β a r e t h e r o o t s o f$ $e q u a t i o n \sqrt{\frac{t}{1 - t}} + \sqrt{\frac{1 - t}{t}} = \frac{13}{6} .$ $F i n d 2 α + 3 β .$
Commented by Rasheed.Sindhi last updated on 14/Nov/21

$D o y o u m e a n \sqrt{\frac{t}{1 - t}} + \sqrt{\frac{1 - t}{t}} = \frac{13}{6}$ $b e c a u s e t h e g i v e n e q u a t i o n l e a d s$ $t o l i n e a r e q u a t i o n w h i c h h a s s i n g l e$ $r o o t α (s a y) o n l y .$
Commented by cortano last updated on 14/Nov/21

$y e s . s o r r y t y p o$
	Answered by Rasheed.Sindhi last updated on 14/Nov/21

	$\sqrt{\frac{t}{1 - t}} + \sqrt{\frac{1 - t}{t}} = \frac{13}{6}$ $\frac{t + (1 - t)}{\sqrt{t (1 - t)}} = \frac{13}{6}$ $\frac{1}{t (1 - t)} = \frac{169}{36}$ $169 t - 169 t^{2} = 36$ $169 t^{2} - 169 t + 36 = 0$ $t = \frac{9}{13}, \frac{4}{13}$ $2 α + 3 β = 2 (\frac{9}{13}) + 3 (\frac{4}{13}), 2 (\frac{4}{13}) + 3 (\frac{9}{13})$ $= \frac{30}{13}, \frac{35}{13}$
	Answered by mr W last updated on 14/Nov/21

	$x = \sqrt{\frac{t}{1 - t}}$ $x + \frac{1}{x} = \frac{13}{6}$ $x^{2} - \frac{13}{6} x + 1 = 0$ $x_{1} x_{2} = 1$ $\Rightarrow \sqrt{\frac{t_{1}}{1 - t_{1}} \times \frac{t_{2}}{1 - t_{2}}} = 1$ $\Rightarrow \sqrt{\frac{α}{1 - α} \times \frac{β}{1 - β}} = 1$ $\Rightarrow α β = 1 - (α + β) + α β$ $\Rightarrow α + β = 1$ $x_{1} + x_{2} = \frac{13}{6}$ $\Rightarrow \sqrt{\frac{t_{1}}{1 - t_{1}}} + \sqrt{\frac{t_{2}}{1 - t_{2}}} = \frac{13}{6}$ $\Rightarrow \sqrt{\frac{α}{1 - α}} + \sqrt{\frac{β}{1 - β}} = \frac{13}{6}$ $\Rightarrow \sqrt{\frac{α}{β}} + \sqrt{\frac{β}{α}} = \frac{13}{6}$ $\Rightarrow \frac{α}{β} + \frac{β}{α} = \frac{97}{36}$ $\Rightarrow \frac{α^{2} + β^{2}}{α β} = \frac{97}{36}$ $\Rightarrow \frac{1 - 2 α β}{α β} = \frac{97}{36}$ $\Rightarrow 169 α β = 36$ $\Rightarrow α β = \frac{36}{169}$ $α, β a r e r o o t s o f z^{2} - z + \frac{36}{169} = 0$ $α, β = z = \frac{1}{2} (1 \pm \frac{5}{13}) = \frac{9}{13}, \frac{4}{13}$ $2 α + 3 β = 2 (α + β) + β = 2 + β$ $= 2 + \frac{9}{13} = \frac{35}{13} o r 2 + \frac{4}{13} = \frac{30}{13}$
	Commented by Rasheed.Sindhi last updated on 14/Nov/21

	$S i r d i d y o u a s s u m e :$ $\sqrt{\frac{t}{1 - t}} + \sqrt{\frac{1 - t}{t}} = \frac{13}{6} ?$
	Commented by mr W last updated on 14/Nov/21

	$y e s . o t h e r w i s e n o n - s e n s e .$
	Commented by Rasheed.Sindhi last updated on 14/Nov/21

	$Your way is tricky one sir!$ $A n y w a y m y w a y s a l s o l e a d t o R o m e . :)$ $R e g a r d l e s s o f d e l a y, {w e}^{'} r e a t l a s t$ $t o g e t h e r i n R o m e! :)$
	Answered by Rasheed.Sindhi last updated on 14/Nov/21
	$(√(t/(1−t))) _(y) + (√((1−t)/t)) = ((13)/6) y+(1/y)=((13)/6) 6y^2 −13y+6=0 (3y−2)(2y−3)=0 y=(2/3) ∨ y=(3/2) (√(t/(1−t)))=(2/3) ∨ (√(t/(1−t)))=(3/2) (t/(1−t))=(4/9) ∨ (t/(1−t))=(9/4) 9t=4−4t ∨ 4t=9−9t t=(4/(13)) ∨ t=(9/(13)) {α,β}={(4/(13)),(9/(13))} 2α+3β=2((4/(13)))+3((9/(13))) , 2((9/(13)))+3((4/(13))) =((35)/(13)),((30)/(13))$
	$\underset{y}{\underset{⏟}{\sqrt{\frac{t}{1 - t}}}} + \sqrt{\frac{1 - t}{t}} = \frac{13}{6}$ $y + \frac{1}{y} = \frac{13}{6}$ $6 y^{2} - 13 y + 6 = 0$ $(3 y - 2) (2 y - 3) = 0$ $y = \frac{2}{3} \lor y = \frac{3}{2}$ $\sqrt{\frac{t}{1 - t}} = \frac{2}{3} \lor \sqrt{\frac{t}{1 - t}} = \frac{3}{2}$ $\frac{t}{1 - t} = \frac{4}{9} \lor \frac{t}{1 - t} = \frac{9}{4}$ $9 t = 4 - 4 t \lor 4 t = 9 - 9 t$ $t = \frac{4}{13} \lor t = \frac{9}{13}$ ${α, β} = {\frac{4}{13}, \frac{9}{13}}$ $2 α + 3 β = 2 (\frac{4}{13}) + 3 (\frac{9}{13}), 2 (\frac{9}{13}) + 3 (\frac{4}{13})$ $= \frac{35}{13}, \frac{30}{13}$
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