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Question Number 159188 by cortano last updated on 14/Nov/21
If α and β are the roots of equation (√(t/(1−t))) + (√((1−t)/t)) = ((13)/6) . Find 2α+3β .
$$\:{If}\:\alpha\:{and}\:\beta\:{are}\:{the}\:{roots}\:{of} \\ $$$${equation}\:\sqrt{\frac{{t}}{\mathrm{1}−{t}}}\:+\:\sqrt{\frac{\mathrm{1}−{t}}{{t}}}\:=\:\frac{\mathrm{13}}{\mathrm{6}}\:. \\ $$$${Find}\:\mathrm{2}\alpha+\mathrm{3}\beta\:. \\ $$
Commented by Rasheed.Sindhi last updated on 14/Nov/21
Do you mean (√(t/(1−t))) + (√((1−t)/t)) = ((13)/6) because the given equation leads to linear equation which has single root α (say) only.
$${Do}\:{you}\:{mean}\:\:\sqrt{\frac{{t}}{\mathrm{1}−{t}}}\:+\:\sqrt{\frac{\mathrm{1}−{t}}{{t}}}\:=\:\frac{\mathrm{13}}{\mathrm{6}} \\ $$$${because}\:{the}\:{given}\:{equation}\:{leads} \\ $$$${to}\:{linear}\:{equation}\:{which}\:{has}\:{single} \\ $$$${root}\:\alpha\:\left({say}\right)\:{only}. \\ $$
Commented by cortano last updated on 14/Nov/21
yes. sorry typo
$${yes}.\:{sorry}\:{typo} \\ $$
Answered by Rasheed.Sindhi last updated on 14/Nov/21
(√(t/(1−t))) + (√((1−t)/t)) = ((13)/6) ((t+(1−t))/( (√(t(1−t)))))=((13)/6) (1/(t(1−t)))=((169)/(36)) 169t−169t^2 =36 169t^2 −169t+36=0 t=(9/(13)),(4/(13)) 2α+3β=2((9/(13)))+3((4/(13))),2((4/(13)))+3((9/(13))) =((30)/(13)),((35)/(13))
$$\sqrt{\frac{{t}}{\mathrm{1}−{t}}}\:+\:\sqrt{\frac{\mathrm{1}−{t}}{{t}}}\:=\:\frac{\mathrm{13}}{\mathrm{6}}\: \\ $$$$\frac{{t}+\left(\mathrm{1}−{t}\right)}{\:\sqrt{{t}\left(\mathrm{1}−{t}\right)}}=\frac{\mathrm{13}}{\mathrm{6}} \\ $$$$\frac{\mathrm{1}}{{t}\left(\mathrm{1}−{t}\right)}=\frac{\mathrm{169}}{\mathrm{36}} \\ $$$$\mathrm{169}{t}−\mathrm{169}{t}^{\mathrm{2}} =\mathrm{36} \\ $$$$\mathrm{169}{t}^{\mathrm{2}} −\mathrm{169}{t}+\mathrm{36}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{9}}{\mathrm{13}},\frac{\mathrm{4}}{\mathrm{13}} \\ $$$$\mathrm{2}\alpha+\mathrm{3}\beta=\mathrm{2}\left(\frac{\mathrm{9}}{\mathrm{13}}\right)+\mathrm{3}\left(\frac{\mathrm{4}}{\mathrm{13}}\right),\mathrm{2}\left(\frac{\mathrm{4}}{\mathrm{13}}\right)+\mathrm{3}\left(\frac{\mathrm{9}}{\mathrm{13}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{30}}{\mathrm{13}},\frac{\mathrm{35}}{\mathrm{13}} \\ $$
Answered by mr W last updated on 14/Nov/21
x=(√(t/(1−t))) x+(1/x)=((13)/6) x^2 −((13)/6)x+1=0 x_1 x_2 =1 ⇒(√((t_1 /(1−t_1 ))×(t_2 /(1−t_2 ))))=1 ⇒(√((α/(1−α))×(β/(1−β))))=1 ⇒αβ=1−(α+β)+αβ ⇒α+β=1 x_1 +x_2 =((13)/6) ⇒(√(t_1 /(1−t_1 )))+(√(t_2 /(1−t_2 )))=((13)/6) ⇒(√(α/(1−α)))+(√(β/(1−β)))=((13)/6) ⇒(√(α/β))+(√(β/α))=((13)/6) ⇒(α/β)+(β/α)=((97)/(36)) ⇒((α^2 +β^2 )/(αβ))=((97)/(36)) ⇒((1−2αβ)/(αβ))=((97)/(36)) ⇒169αβ=36 ⇒αβ=((36)/(169)) α,β are roots of z^2 −z+((36)/(169))=0 α,β=z=(1/2)(1±(5/(13)))=(9/(13)), (4/(13)) 2α+3β=2(α+β)+β=2+β =2+(9/(13))=((35)/(13)) or 2+(4/(13))=((30)/(13))
$${x}=\sqrt{\frac{{t}}{\mathrm{1}−{t}}} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\frac{\mathrm{13}}{\mathrm{6}} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{13}}{\mathrm{6}}{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} =\mathrm{1}\: \\ $$$$\Rightarrow\sqrt{\frac{{t}_{\mathrm{1}} }{\mathrm{1}−{t}_{\mathrm{1}} }×\frac{{t}_{\mathrm{2}} }{\mathrm{1}−{t}_{\mathrm{2}} }}=\mathrm{1} \\ $$$$\Rightarrow\sqrt{\frac{\alpha}{\mathrm{1}−\alpha}×\frac{\beta}{\mathrm{1}−\beta}}=\mathrm{1} \\ $$$$\Rightarrow\alpha\beta=\mathrm{1}−\left(\alpha+\beta\right)+\alpha\beta \\ $$$$\Rightarrow\alpha+\beta=\mathrm{1} \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} =\frac{\mathrm{13}}{\mathrm{6}} \\ $$$$\Rightarrow\sqrt{\frac{{t}_{\mathrm{1}} }{\mathrm{1}−{t}_{\mathrm{1}} }}+\sqrt{\frac{{t}_{\mathrm{2}} }{\mathrm{1}−{t}_{\mathrm{2}} }}=\frac{\mathrm{13}}{\mathrm{6}} \\ $$$$\Rightarrow\sqrt{\frac{\alpha}{\mathrm{1}−\alpha}}+\sqrt{\frac{\beta}{\mathrm{1}−\beta}}=\frac{\mathrm{13}}{\mathrm{6}} \\ $$$$\Rightarrow\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}=\frac{\mathrm{13}}{\mathrm{6}} \\ $$$$\Rightarrow\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\mathrm{97}}{\mathrm{36}} \\ $$$$\Rightarrow\frac{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }{\alpha\beta}=\frac{\mathrm{97}}{\mathrm{36}} \\ $$$$\Rightarrow\frac{\mathrm{1}−\mathrm{2}\alpha\beta}{\alpha\beta}=\frac{\mathrm{97}}{\mathrm{36}} \\ $$$$\Rightarrow\mathrm{169}\alpha\beta=\mathrm{36} \\ $$$$\Rightarrow\alpha\beta=\frac{\mathrm{36}}{\mathrm{169}} \\ $$$$\alpha,\beta\:{are}\:{roots}\:{of}\:{z}^{\mathrm{2}} −{z}+\frac{\mathrm{36}}{\mathrm{169}}=\mathrm{0} \\ $$$$\alpha,\beta={z}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\pm\frac{\mathrm{5}}{\mathrm{13}}\right)=\frac{\mathrm{9}}{\mathrm{13}},\:\frac{\mathrm{4}}{\mathrm{13}} \\ $$$$ \\ $$$$\mathrm{2}\alpha+\mathrm{3}\beta=\mathrm{2}\left(\alpha+\beta\right)+\beta=\mathrm{2}+\beta \\ $$$$=\mathrm{2}+\frac{\mathrm{9}}{\mathrm{13}}=\frac{\mathrm{35}}{\mathrm{13}}\:{or}\:\mathrm{2}+\frac{\mathrm{4}}{\mathrm{13}}=\frac{\mathrm{30}}{\mathrm{13}} \\ $$
Commented by Rasheed.Sindhi last updated on 14/Nov/21
Sir did you assume: (√(t/(1−t))) + (√((1−t)/t)) = ((13)/6) ?
$$\mathcal{S}{ir}\:{did}\:{you}\:{assume}: \\ $$$$\:\sqrt{\frac{{t}}{\mathrm{1}−{t}}}\:+\:\sqrt{\frac{\mathrm{1}−{t}}{{t}}}\:=\:\frac{\mathrm{13}}{\mathrm{6}}\:\:? \\ $$
Commented by mr W last updated on 14/Nov/21
yes. otherwise non−sense.
$${yes}.\:{otherwise}\:{non}−{sense}. \\ $$
Commented by Rasheed.Sindhi last updated on 14/Nov/21
Your way is tricky one sir! Anyway my ways also lead to Rome.:) Regardless of delay, we′re at last together in Rome! :)
$$\mathrm{Your}\:\mathrm{way}\:\mathrm{is}\:\mathrm{tricky}\:\mathrm{one}\:\:\mathrm{sir}! \\ $$$$\left.{Anyway}\:{my}\:{ways}\:{also}\:{lead}\:{to}\:{Rome}.:\right) \\ $$$${Regardless}\:{of}\:{delay},\:{we}'{re}\:{at}\:{last} \\ $$$$\left.{together}\:{in}\:{Rome}!\::\right) \\ $$
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))
$$\:\underset{{y}} {\underbrace{\sqrt{\frac{{t}}{\mathrm{1}−{t}}}\:}}+\:\sqrt{\frac{\mathrm{1}−{t}}{{t}}}\:=\:\frac{\mathrm{13}}{\mathrm{6}} \\ $$$${y}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{13}}{\mathrm{6}} \\ $$$$\mathrm{6}{y}^{\mathrm{2}} −\mathrm{13}{y}+\mathrm{6}=\mathrm{0} \\ $$$$\left(\mathrm{3}{y}−\mathrm{2}\right)\left(\mathrm{2}{y}−\mathrm{3}\right)=\mathrm{0} \\ $$$${y}=\frac{\mathrm{2}}{\mathrm{3}}\:\vee\:{y}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\sqrt{\frac{{t}}{\mathrm{1}−{t}}}=\frac{\mathrm{2}}{\mathrm{3}}\:\vee\:\sqrt{\frac{{t}}{\mathrm{1}−{t}}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\frac{{t}}{\mathrm{1}−{t}}=\frac{\mathrm{4}}{\mathrm{9}}\:\vee\:\frac{{t}}{\mathrm{1}−{t}}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\mathrm{9}{t}=\mathrm{4}−\mathrm{4}{t}\:\vee\:\mathrm{4}{t}=\mathrm{9}−\mathrm{9}{t} \\ $$$${t}=\frac{\mathrm{4}}{\mathrm{13}}\:\vee\:{t}=\frac{\mathrm{9}}{\mathrm{13}} \\ $$$$\left\{\alpha,\beta\right\}=\left\{\frac{\mathrm{4}}{\mathrm{13}},\frac{\mathrm{9}}{\mathrm{13}}\right\} \\ $$$$\mathrm{2}\alpha+\mathrm{3}\beta=\mathrm{2}\left(\frac{\mathrm{4}}{\mathrm{13}}\right)+\mathrm{3}\left(\frac{\mathrm{9}}{\mathrm{13}}\right)\:,\:\mathrm{2}\left(\frac{\mathrm{9}}{\mathrm{13}}\right)+\mathrm{3}\left(\frac{\mathrm{4}}{\mathrm{13}}\right) \\ $$$$\:\:\:\:\:=\frac{\mathrm{35}}{\mathrm{13}},\frac{\mathrm{30}}{\mathrm{13}} \\ $$

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