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Question Number 117026 by bemath last updated on 08/Oct/20

((√(1−x))/( (√x))) + ((√x)/( (√(1−x)))) = ((13)/6)

$$\:\frac{\sqrt{\mathrm{1}−\mathrm{x}}}{\:\sqrt{\mathrm{x}}}\:+\:\frac{\sqrt{\mathrm{x}}}{\:\sqrt{\mathrm{1}−\mathrm{x}}}\:=\:\frac{\mathrm{13}}{\mathrm{6}} \\ $$

Answered by 1549442205PVT last updated on 09/Oct/20

We need the condition for the equation is defined as 0<x<1(∗) ((√(1−x))/( (√x))) + ((√x)/( (√(1−x)))) = ((13)/6) (1) (1)⇔((1−x+x)/( (√(x(1−x)))))=((13)/6) ⇔6=13(√(x(1−x)))⇔36=169(x−x^2 ) ⇔169x^2 −169x+36=0 Δ=169^2 −4.169.36=169.25=65^2 ⇒x=((169±65)/(2.169))=((13±5)/(2.13))∈{(4/(13)),(9/(13))} both satisfy (∗) Thus,the given equation has two roots x∈{(4/(13)),(9/(13))}

$$\mathrm{We}\:\mathrm{need}\:\mathrm{the}\:\mathrm{condition}\:\mathrm{for}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{is}\:\mathrm{defined}\:\mathrm{as}\:\mathrm{0}<\mathrm{x}<\mathrm{1}\left(\ast\right) \\ $$$$\:\frac{\sqrt{\mathrm{1}−\mathrm{x}}}{\:\sqrt{\mathrm{x}}}\:+\:\frac{\sqrt{\mathrm{x}}}{\:\sqrt{\mathrm{1}−\mathrm{x}}}\:=\:\frac{\mathrm{13}}{\mathrm{6}}\:\left(\mathrm{1}\right) \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\frac{\mathrm{1}−\mathrm{x}+\mathrm{x}}{\:\sqrt{\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)}}=\frac{\mathrm{13}}{\mathrm{6}} \\ $$$$\Leftrightarrow\mathrm{6}=\mathrm{13}\sqrt{\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)}\Leftrightarrow\mathrm{36}=\mathrm{169}\left(\mathrm{x}−\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\Leftrightarrow\mathrm{169x}^{\mathrm{2}} −\mathrm{169x}+\mathrm{36}=\mathrm{0} \\ $$$$\Delta=\mathrm{169}^{\mathrm{2}} −\mathrm{4}.\mathrm{169}.\mathrm{36}=\mathrm{169}.\mathrm{25}=\mathrm{65}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{169}\pm\mathrm{65}}{\mathrm{2}.\mathrm{169}}=\frac{\mathrm{13}\pm\mathrm{5}}{\mathrm{2}.\mathrm{13}}\in\left\{\frac{\mathrm{4}}{\mathrm{13}},\frac{\mathrm{9}}{\mathrm{13}}\right\} \\ $$$$\mathrm{both}\:\mathrm{satisfy}\:\left(\ast\right) \\ $$$$\mathrm{Thus},\mathrm{the}\:\mathrm{given}\:\mathrm{equation}\:\mathrm{has}\:\mathrm{two}\:\mathrm{roots} \\ $$$$\mathrm{x}\in\left\{\frac{\mathrm{4}}{\mathrm{13}},\frac{\mathrm{9}}{\mathrm{13}}\right\} \\ $$

Commented by bemath last updated on 09/Oct/20

gave kudos

$$\mathrm{gave}\:\mathrm{kudos} \\ $$

Answered by TANMAY PANACEA last updated on 09/Oct/20

a+(1/a)=((13)/6)→6a^2 −13a+6=0 6a^2 −9a−4a+6=0 3a(2a−3)−2(2a−3)=0 (2a−3)(3a−2)=0 a=(3/2),(2/3) a^2 =((1−x)/x)=(9/4) and ((1−x)/x)=(4/9) 9x+4x=4→x=(4/(13)) and 13x=9 x=(9/(13))

$${a}+\frac{\mathrm{1}}{{a}}=\frac{\mathrm{13}}{\mathrm{6}}\rightarrow\mathrm{6}{a}^{\mathrm{2}} −\mathrm{13}{a}+\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{6}{a}^{\mathrm{2}} −\mathrm{9}{a}−\mathrm{4}{a}+\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{3}{a}\left(\mathrm{2}{a}−\mathrm{3}\right)−\mathrm{2}\left(\mathrm{2}{a}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\left(\mathrm{2}{a}−\mathrm{3}\right)\left(\mathrm{3}{a}−\mathrm{2}\right)=\mathrm{0} \\ $$$${a}=\frac{\mathrm{3}}{\mathrm{2}},\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{1}−{x}}{{x}}=\frac{\mathrm{9}}{\mathrm{4}}\:\:\:\:{and}\:\:\frac{\mathrm{1}−{x}}{{x}}=\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\mathrm{9}{x}+\mathrm{4}{x}=\mathrm{4}\rightarrow\boldsymbol{{x}}=\frac{\mathrm{4}}{\mathrm{13}}\:\boldsymbol{{and}}\:\mathrm{13}\boldsymbol{{x}}=\mathrm{9}\:\:\:\:\boldsymbol{{x}}=\frac{\mathrm{9}}{\mathrm{13}} \\ $$

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