If-and-are-the-roots-of-equation-t-1-t-1-t-t-13-6-Find-2-3-

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

\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}