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x-1-y-2-y-2-x-1-1-gt-x-




Question Number 111008 by Khanacademy last updated on 01/Sep/20
((√(x+1))/(y+2)) + ((√(y+2))/(x+1)) =1      =>  x=?
$$\frac{\sqrt{\boldsymbol{{x}}+\mathrm{1}}}{\boldsymbol{{y}}+\mathrm{2}}\:+\:\frac{\sqrt{\boldsymbol{{y}}+\mathrm{2}}}{\boldsymbol{{x}}+\mathrm{1}}\:=\mathrm{1}\:\:\:\:\:\:=>\:\:\boldsymbol{{x}}=? \\ $$
Commented by Rasheed.Sindhi last updated on 01/Sep/20
x in terms of y?
$${x}\:{in}\:{terms}\:{of}\:{y}? \\ $$
Commented by Khanacademy last updated on 01/Sep/20
???????
$$??????? \\ $$
Commented by Her_Majesty last updated on 01/Sep/20
complicated...  (a/b^2 )+(b/a^2 )=1  a^3 −a^2 b^2 +b^3 =0  a=t+(b^2 /3)  t^3 −(b^4 /3)t−((b^3 (2b^3 −27))/(27))=0  we can solve this for t  D=(b^6 /4)−(b^9 /(27))  we need Cardano′s or the Trigonometric  solution  depending on the desired domain:  R: x>−1∧y>−2 ⇒ a>0∧b>0  C: x≠−1∧y≠−2 ⇒ a≠0∧b≠0
$${complicated}… \\ $$$$\frac{{a}}{{b}^{\mathrm{2}} }+\frac{{b}}{{a}^{\mathrm{2}} }=\mathrm{1} \\ $$$${a}^{\mathrm{3}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{3}} =\mathrm{0} \\ $$$${a}={t}+\frac{{b}^{\mathrm{2}} }{\mathrm{3}} \\ $$$${t}^{\mathrm{3}} −\frac{{b}^{\mathrm{4}} }{\mathrm{3}}{t}−\frac{{b}^{\mathrm{3}} \left(\mathrm{2}{b}^{\mathrm{3}} −\mathrm{27}\right)}{\mathrm{27}}=\mathrm{0} \\ $$$${we}\:{can}\:{solve}\:{this}\:{for}\:{t} \\ $$$${D}=\frac{{b}^{\mathrm{6}} }{\mathrm{4}}−\frac{{b}^{\mathrm{9}} }{\mathrm{27}} \\ $$$${we}\:{need}\:{Cardano}'{s}\:{or}\:{the}\:{Trigonometric} \\ $$$${solution} \\ $$$${depending}\:{on}\:{the}\:{desired}\:{domain}: \\ $$$$\mathbb{R}:\:{x}>−\mathrm{1}\wedge{y}>−\mathrm{2}\:\Rightarrow\:{a}>\mathrm{0}\wedge{b}>\mathrm{0} \\ $$$$\mathbb{C}:\:{x}\neq−\mathrm{1}\wedge{y}\neq−\mathrm{2}\:\Rightarrow\:{a}\neq\mathrm{0}\wedge{b}\neq\mathrm{0} \\ $$

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