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x-2-x-c-find-all-possible-solutions-for-x-C-and-1-c-0-2-c-R-3-c-C-

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Question Number 228118 by Ghisom_ last updated on 25/Mar/26
x^2 +(√x)=c find all possible solutions for x∈C and 1. c≥0 2. c∈R 3. c∈C
$${x}^{\mathrm{2}} +\sqrt{{x}}={c} \\ $$$$\mathrm{find}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{solutions}\:\mathrm{for}\:{x}\in\mathbb{C}\:\mathrm{and} \\ $$$$\mathrm{1}.\:{c}\geqslant\mathrm{0} \\ $$$$\mathrm{2}.\:{c}\in\mathbb{R} \\ $$$$\mathrm{3}.\:{c}\in\mathbb{C} \\ $$
Answered by ajfour last updated on 31/Mar/26
let (√x)=t t^4 +t=c (t^2 +pt+h)(t^2 −pt+k)=0 h+k=p^2 p(k−h)=1 hk=−c p^4 −(1/p^2 )+4c=0 p^6 +4cp^2 −1=0 p^2 =((√((1/4)+(((4c)/3))^3 ))+(1/2))^(1/3) −((√((1/4)+(((4c)/3))^3 ))−(1/2))^(1/3) 2h=p^2 −(1/p) t=−(p/2)±(√((p^2 /4)+(1/(2p))−(p^2 /2))) x=t^2 =−pt−h x=(1/(2p))∓(√((p/2)−(p^4 /4)))
$${let}\:\sqrt{{x}}={t} \\ $$$${t}^{\mathrm{4}} +{t}={c} \\ $$$$\left({t}^{\mathrm{2}} +{pt}+{h}\right)\left({t}^{\mathrm{2}} −{pt}+{k}\right)=\mathrm{0} \\ $$$${h}+{k}={p}^{\mathrm{2}} \\ $$$${p}\left({k}−{h}\right)=\mathrm{1} \\ $$$${hk}=−{c} \\ $$$${p}^{\mathrm{4}} −\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\mathrm{4}{c}=\mathrm{0} \\ $$$${p}^{\mathrm{6}} +\mathrm{4}{cp}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${p}^{\mathrm{2}} =\left(\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\left(\frac{\mathrm{4}{c}}{\mathrm{3}}\right)^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:−\left(\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\left(\frac{\mathrm{4}{c}}{\mathrm{3}}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\mathrm{2}{h}={p}^{\mathrm{2}} −\frac{\mathrm{1}}{{p}} \\ $$$${t}=−\frac{{p}}{\mathrm{2}}\pm\sqrt{\frac{{p}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}{p}}−\frac{{p}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$${x}={t}^{\mathrm{2}} =−{pt}−{h} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}{p}}\mp\sqrt{\frac{{p}}{\mathrm{2}}−\frac{{p}^{\mathrm{4}} }{\mathrm{4}}} \\ $$
Answered by Jyrgen last updated on 03/Apr/26
c=a^2 +(√a); a∈C x^2 +(√x)=a^2 +(√a) x^2 −a^2 =−((√x)−(√a)) ((√x)−(√a))((√x)+(√a))(x+a)=−((√x)−(√a)) obviously x_1 =a ((√x)+(√a))(x+a)=−1 now it′s getting complicated...
$${c}={a}^{\mathrm{2}} +\sqrt{{a}};\:{a}\in\mathbb{C} \\ $$$${x}^{\mathrm{2}} +\sqrt{{x}}={a}^{\mathrm{2}} +\sqrt{{a}} \\ $$$${x}^{\mathrm{2}} −{a}^{\mathrm{2}} =−\left(\sqrt{{x}}−\sqrt{{a}}\right) \\ $$$$\left(\sqrt{{x}}−\sqrt{{a}}\right)\left(\sqrt{{x}}+\sqrt{{a}}\right)\left({x}+{a}\right)=−\left(\sqrt{{x}}−\sqrt{{a}}\right) \\ $$$${obviously}\:{x}_{\mathrm{1}} ={a} \\ $$$$\left(\sqrt{{x}}+\sqrt{{a}}\right)\left({x}+{a}\right)=−\mathrm{1} \\ $$$${now}\:{it}'{s}\:{getting}\:{complicated}… \\ $$
Answered by Frix last updated on 05/Apr/26
For c∈R and x∈C: u, v, w ∈R Because of symmetry we can let w≥0 but for (√x)=u+wi we need u≥0 or else x doesn′t exist {: (((√x)=u+wi)),((x^2 =v−wi)) } ⇒ x^2 +(√x)=u+v (√x)=u+wi ⇒ x^2 =(u+wi)^4 (u+wi)^4 =v−wi ⇔ { ((u^4 +w^4 −6u^2 w^2 −v=0)),(((4u^3 −4uw^2 +1)w=0)) :} 1. w=0 ⇒ v=u^4 and we end up finding the real solution for x^2 +(√x)=u^4 +u=c≥0 2. w≠0 { ((u^4 +w^4 −6u^2 w^2 −v=0)),((4u^3 −4uw^2 +1=0 ⇒ w=((√(4u^3 +1))/(2(√u))))) :} { ((v=−4u^4 −u+(1/(16u^2 )))),((w=((√(4u^3 +1))/(2(√u))))) :} u+v=c ⇔ v=c−u ⇒ We have to solve u^6 +(c/4)u^2 −(1/(64))=0 for u∈R. It is possible to solve exactly (but not useful in most cases) using u=(√t) ⇒ t^3 +(c/4)t−(1/(64))=0; t≥0 x^2 +(√x)=c≥0 ⇒ 1 real and 2 conjugated complex solutions x^2 +(√x)=<0 ⇒ 2 conjugated complex solutions
$$\mathrm{For}\:{c}\in\mathbb{R}\:\mathrm{and}\:{x}\in\mathbb{C}: \\ $$$${u},\:{v},\:{w}\:\in\mathbb{R} \\ $$$$\mathrm{Because}\:\mathrm{of}\:\mathrm{symmetry}\:\mathrm{we}\:\mathrm{can}\:\mathrm{let}\:{w}\geqslant\mathrm{0}\:\mathrm{but} \\ $$$$\mathrm{for}\:\sqrt{{x}}={u}+{w}\mathrm{i}\:\mathrm{we}\:\mathrm{need}\:{u}\geqslant\mathrm{0}\:\mathrm{or}\:\mathrm{else}\:{x}\:\mathrm{doesn}'\mathrm{t} \\ $$$$\mathrm{exist} \\ $$$$\left.\begin{matrix}{\sqrt{{x}}={u}+{w}\mathrm{i}}\\{{x}^{\mathrm{2}} ={v}−{w}\mathrm{i}}\end{matrix}\right\}\:\Rightarrow\:{x}^{\mathrm{2}} +\sqrt{{x}}={u}+{v} \\ $$$$\sqrt{{x}}={u}+{w}\mathrm{i}\:\Rightarrow\:{x}^{\mathrm{2}} =\left({u}+{w}\mathrm{i}\right)^{\mathrm{4}} \\ $$$$\left({u}+{w}\mathrm{i}\right)^{\mathrm{4}} ={v}−{w}\mathrm{i}\:\Leftrightarrow\:\begin{cases}{{u}^{\mathrm{4}} +{w}^{\mathrm{4}} −\mathrm{6}{u}^{\mathrm{2}} {w}^{\mathrm{2}} −{v}=\mathrm{0}}\\{\left(\mathrm{4}{u}^{\mathrm{3}} −\mathrm{4}{uw}^{\mathrm{2}} +\mathrm{1}\right){w}=\mathrm{0}}\end{cases} \\ $$$$\mathrm{1}.\:{w}=\mathrm{0}\:\Rightarrow\:{v}={u}^{\mathrm{4}} \:\mathrm{and}\:\mathrm{we}\:\mathrm{end}\:\mathrm{up}\:\mathrm{finding}\:\mathrm{the} \\ $$$$\mathrm{real}\:\mathrm{solution}\:\mathrm{for}\:{x}^{\mathrm{2}} +\sqrt{{x}}={u}^{\mathrm{4}} +{u}={c}\geqslant\mathrm{0} \\ $$$$\mathrm{2}.\:{w}\neq\mathrm{0} \\ $$$$\begin{cases}{{u}^{\mathrm{4}} +{w}^{\mathrm{4}} −\mathrm{6}{u}^{\mathrm{2}} {w}^{\mathrm{2}} −{v}=\mathrm{0}}\\{\mathrm{4}{u}^{\mathrm{3}} −\mathrm{4}{uw}^{\mathrm{2}} +\mathrm{1}=\mathrm{0}\:\Rightarrow\:{w}=\frac{\sqrt{\mathrm{4}{u}^{\mathrm{3}} +\mathrm{1}}}{\mathrm{2}\sqrt{{u}}}}\end{cases} \\ $$$$\begin{cases}{{v}=−\mathrm{4}{u}^{\mathrm{4}} −{u}+\frac{\mathrm{1}}{\mathrm{16}{u}^{\mathrm{2}} }}\\{{w}=\frac{\sqrt{\mathrm{4}{u}^{\mathrm{3}} +\mathrm{1}}}{\mathrm{2}\sqrt{{u}}}}\end{cases} \\ $$$${u}+{v}={c}\:\Leftrightarrow\:{v}={c}−{u} \\ $$$$\Rightarrow\:\mathrm{We}\:\mathrm{have}\:\mathrm{to}\:\mathrm{solve} \\ $$$${u}^{\mathrm{6}} +\frac{{c}}{\mathrm{4}}{u}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{64}}=\mathrm{0} \\ $$$$\mathrm{for}\:{u}\in\mathbb{R}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{exactly}\:\left(\mathrm{but}\right. \\ $$$$\left.\mathrm{not}\:\mathrm{useful}\:\mathrm{in}\:\mathrm{most}\:\mathrm{cases}\right)\:\mathrm{using}\:{u}=\sqrt{{t}}\:\Rightarrow \\ $$$${t}^{\mathrm{3}} +\frac{{c}}{\mathrm{4}}{t}−\frac{\mathrm{1}}{\mathrm{64}}=\mathrm{0};\:{t}\geqslant\mathrm{0} \\ $$$$ \\ $$$${x}^{\mathrm{2}} +\sqrt{{x}}={c}\geqslant\mathrm{0}\:\Rightarrow\:\mathrm{1}\:\mathrm{real}\:\mathrm{and}\:\mathrm{2}\:\mathrm{conjugated} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{complex}\:\mathrm{solutions} \\ $$$${x}^{\mathrm{2}} +\sqrt{{x}}=<\mathrm{0}\:\Rightarrow\:\mathrm{2}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{solutions} \\ $$

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