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Question Number 90194 by behi83417@gmail.com last updated on 21/Apr/20
(x^2 /(x+a))+(√x)=a (a∈R) solve for: x .
$$\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{a}}}+\sqrt{\boldsymbol{\mathrm{x}}}=\boldsymbol{\mathrm{a}}\:\:\:\:\:\left(\boldsymbol{\mathrm{a}}\in\boldsymbol{\mathrm{R}}\right) \\ $$$$\mathrm{solve}\:\mathrm{for}:\:\:\mathrm{x}\:\:. \\ $$
Answered by ajfour last updated on 22/Apr/20
let (√x)=t t^4 =(a−t)(t^2 +a) t^4 +t^3 −at^2 +at−a^2 =0 let (t^2 +pt+q)(t^2 +rt+m)=0 ⇒ t^4 +(p+r)t^3 +(m+pr+q)t^2 +(pm+qr)t+qm=0 ⇒ p+r=1 , pr+m+q=−a, pm+qr =a , qm=−a^2 ⇒ pm−((a^2 r)/m)=a pm^2 −am−a^2 r=0 m=(a/(2p))±(√((a^2 /(4p^2 ))+((a^2 r)/p))) m= (a/(2p)){1±(√(1+4pr))} q=−(a^2 /m) = (2ap)(({1∓(√(1+4pr))})/(4pr)) q= (a/(2r)){1∓(√(1+4pr))} q+m=(a/(2p))+(a/(2r))±(√(1+4pr))((a/(2p))−(a/(2r))) −a−pr=(a/2)(((p+r)/(pr)))±(a/2)(((r−p)/(pr)))(√(1+4pr)) let pr=z −a−z=(a/(2z))±(a/(2z))(√(1−16z^2 )) ⇒ ((a/(2z))+z+a)^2 =(a^2 /(4z^2 ))(1−16z^2 ) z^2 +a^2 +a+2az+(a^2 /z)+4a^2 =0 z^3 +2az^2 +a(5a+1)z+a^2 =0 let z=y−((2a)/3) y^3 −2ay^2 +((4a^2 y)/3)−((8a^3 )/(27)) +2ay^2 −((8a^2 y)/3)+((8a^3 )/9) +5a^2 y+ay−((10a^3 )/3)−((2a^2 )/3)+a^2 =0 y^3 +(((11a^2 )/3)+a)y−((70a^3 )/(27))+(a^2 /3)=0 .........
$${let}\:\sqrt{{x}}={t} \\ $$$${t}^{\mathrm{4}} =\left({a}−{t}\right)\left({t}^{\mathrm{2}} +{a}\right) \\ $$$${t}^{\mathrm{4}} +{t}^{\mathrm{3}} −{at}^{\mathrm{2}} +{at}−{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${let} \\ $$$$\left({t}^{\mathrm{2}} +{pt}+{q}\right)\left({t}^{\mathrm{2}} +{rt}+{m}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${t}^{\mathrm{4}} +\left({p}+{r}\right){t}^{\mathrm{3}} +\left({m}+{pr}+{q}\right){t}^{\mathrm{2}} \\ $$$$\:\:+\left({pm}+{qr}\right){t}+{qm}=\mathrm{0} \\ $$$$\Rightarrow\:\:{p}+{r}=\mathrm{1}\:\:,\:\:{pr}+{m}+{q}=−{a}, \\ $$$$\:\:{pm}+{qr}\:={a}\:,\:\:{qm}=−{a}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{pm}−\frac{{a}^{\mathrm{2}} {r}}{{m}}={a} \\ $$$${pm}^{\mathrm{2}} −{am}−{a}^{\mathrm{2}} {r}=\mathrm{0} \\ $$$${m}=\frac{{a}}{\mathrm{2}{p}}\pm\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}{p}^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} {r}}{{p}}} \\ $$$$\:\:\:{m}=\:\frac{{a}}{\mathrm{2}{p}}\left\{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}{pr}}\right\} \\ $$$$\:\:{q}=−\frac{{a}^{\mathrm{2}} }{{m}}\:=\:\left(\mathrm{2}{ap}\right)\frac{\left\{\mathrm{1}\mp\sqrt{\mathrm{1}+\mathrm{4}{pr}}\right\}}{\mathrm{4}{pr}} \\ $$$$\:\:{q}=\:\frac{{a}}{\mathrm{2}{r}}\left\{\mathrm{1}\mp\sqrt{\mathrm{1}+\mathrm{4}{pr}}\right\} \\ $$$${q}+{m}=\frac{{a}}{\mathrm{2}{p}}+\frac{{a}}{\mathrm{2}{r}}\pm\sqrt{\mathrm{1}+\mathrm{4}{pr}}\left(\frac{{a}}{\mathrm{2}{p}}−\frac{{a}}{\mathrm{2}{r}}\right) \\ $$$$−{a}−{pr}=\frac{{a}}{\mathrm{2}}\left(\frac{{p}+{r}}{{pr}}\right)\pm\frac{{a}}{\mathrm{2}}\left(\frac{{r}−{p}}{{pr}}\right)\sqrt{\mathrm{1}+\mathrm{4}{pr}} \\ $$$${let}\:\:{pr}={z} \\ $$$$−{a}−{z}=\frac{{a}}{\mathrm{2}{z}}\pm\frac{{a}}{\mathrm{2}{z}}\sqrt{\mathrm{1}−\mathrm{16}{z}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\left(\frac{{a}}{\mathrm{2}{z}}+{z}+{a}\right)^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{\mathrm{4}{z}^{\mathrm{2}} }\left(\mathrm{1}−\mathrm{16}{z}^{\mathrm{2}} \right) \\ $$$${z}^{\mathrm{2}} +{a}^{\mathrm{2}} +{a}+\mathrm{2}{az}+\frac{{a}^{\mathrm{2}} }{{z}}+\mathrm{4}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${z}^{\mathrm{3}} +\mathrm{2}{az}^{\mathrm{2}} +{a}\left(\mathrm{5}{a}+\mathrm{1}\right){z}+{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${let}\:\:{z}={y}−\frac{\mathrm{2}{a}}{\mathrm{3}} \\ $$$${y}^{\mathrm{3}} −\mathrm{2}{ay}^{\mathrm{2}} +\frac{\mathrm{4}{a}^{\mathrm{2}} {y}}{\mathrm{3}}−\frac{\mathrm{8}{a}^{\mathrm{3}} }{\mathrm{27}} \\ $$$$\:\:\:\:+\mathrm{2}{ay}^{\mathrm{2}} −\frac{\mathrm{8}{a}^{\mathrm{2}} {y}}{\mathrm{3}}+\frac{\mathrm{8}{a}^{\mathrm{3}} }{\mathrm{9}} \\ $$$$\:\:\:+\mathrm{5}{a}^{\mathrm{2}} {y}+{ay}−\frac{\mathrm{10}{a}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{3}}+{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${y}^{\mathrm{3}} +\left(\frac{\mathrm{11}{a}^{\mathrm{2}} }{\mathrm{3}}+{a}\right){y}−\frac{\mathrm{70}{a}^{\mathrm{3}} }{\mathrm{27}}+\frac{{a}^{\mathrm{2}} }{\mathrm{3}}=\mathrm{0} \\ $$$$……… \\ $$
Commented by behi83417@gmail.com last updated on 22/Apr/20
thank you very much dear Ajfour.
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{Ajfour}. \\ $$
Commented by ajfour last updated on 22/Apr/20
i will try to finish..
$${i}\:{will}\:{try}\:{to}\:{finish}.. \\ $$

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