Question Number 80144 by behi83417@gmail.com last updated on 31/Jan/20
![solve for x: (((√x)+1)/(√(x+1)))+ax^2 =x(a^2 +1) [a∈R]](Q80144.png)
$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{x}}: \\ $$$$\frac{\sqrt{\boldsymbol{\mathrm{x}}}+\mathrm{1}}{\sqrt{\boldsymbol{\mathrm{x}}+\mathrm{1}}}+\boldsymbol{\mathrm{ax}}^{\mathrm{2}} =\boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\mathrm{1}\right)\:\:\:\:\:\:\left[\boldsymbol{\mathrm{a}}\in\boldsymbol{\mathrm{R}}\right] \\ $$
Commented by john santu last updated on 31/Jan/20
$${let}\:\sqrt{{x}+\mathrm{1}}\:=\:\mathrm{sec}\:{t} \\ $$$${x}+\mathrm{1}\:=\:\mathrm{sec}\:^{\mathrm{2}} {t}\Rightarrow{x}\:=\:\mathrm{tan}\:^{\mathrm{2}} {t} \\ $$$$\frac{\mathrm{tan}\:{t}+\mathrm{1}}{\mathrm{sec}\:{t}}+{a}\left(\mathrm{tan}\:^{\mathrm{4}} {t}\right)=\left({a}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{tan}\:^{\mathrm{2}} {t} \\ $$$$\mathrm{cos}\:{t}\left(\mathrm{tan}\:{t}+\mathrm{1}\right)+{a}\mathrm{tan}\:^{\mathrm{4}} {t}=\left({a}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{tan}\:^{\mathrm{2}} {t} \\ $$