Question Number 221501 by Jyrgen last updated on 07/Jun/25
$${solve}\:{for}\:\mathrm{x}: \\ $$$$\sqrt{\mathrm{a}−\sqrt{\mathrm{a}+\mathrm{x}}}+\sqrt{\mathrm{a}+\sqrt{\mathrm{a}−\mathrm{x}}}=\mathrm{2x} \\ $$$${it}'{s}\:{possible}\:{to}\:{solve}\:{for}\:\mathrm{a}\:{but}\:\mathrm{x}\:{seems} \\ $$$${impossible}\:{to}\:{me} \\ $$
Answered by ajfour last updated on 07/Jun/25
$${x}=\left({Lycon}\:{Trix}\:−\:{Alphaprime}\right)^{{a}} \\ $$
Commented by mr W last updated on 07/Jun/25
$${is}\:{this}\:{that}\:{famous}\:{equation}\:{which} \\ $$$${you}\:{have}\:{solved}? \\ $$
Commented by ajfour last updated on 07/Jun/25
$${yeah}\:{this}\:{was}\:{it}. \\ $$
Commented by Tawa11 last updated on 07/Jun/25
$$\mathrm{Sirs},\:\mathrm{are}\:\mathrm{you}\:\mathrm{joking},\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand} \\ $$$$\mathrm{anything}.\:\mathrm{Is}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{exist}?? \\ $$$$\mathrm{I}\:\mathrm{am}\:\mathrm{asking}\:\mathrm{to}\:\mathrm{learn}. \\ $$
Commented by Jyrgen last updated on 07/Jun/25
$${so}\:{you}\:{found}\:{the}\:{solution}?\:{is}\:{it}\:{here}\:{on} \\ $$$${the}\:{forum}? \\ $$
Commented by mr W last updated on 07/Jun/25
$${ajfour}\:{sir}\:{and}\:{MJS}\:{sir}\:{have}\:{jointly} \\ $$$${solved}\:{this}\:{problem}.\:{it}\:{is}\:{said}\:{that} \\ $$$${prize}\:{money}\:{was}\:{offered}\:{for}\:{the} \\ $$$${solution}\:{of}\:{this}\:{problem}.\: \\ $$$${while}\:{ajfour}\:{sir}\:{is}\:{still}\:{active}\:{in}\: \\ $$$${this}\:{forum},\:{MJS}\:{sir}\:{has}\:{been}\: \\ $$$${missed}\:{here}\:{for}\:{a}\:{long}\:{time}.\:{i}\:{wish}\: \\ $$$${him}\:{all}\:{the}\:{best}. \\ $$$${the}\:{result}\:{of}\:{their}\:{solution}\:{is}: \\ $$
Commented by mr W last updated on 07/Jun/25
Commented by Tawa11 last updated on 07/Jun/25
$$\mathrm{Wow}. \\ $$$$\mathrm{Great}. \\ $$$$\mathrm{Hope}\:\mathrm{they}\:\mathrm{have}\:\mathrm{collected}\:\mathrm{the}\:\mathrm{money} \\ $$
Commented by Tawa11 last updated on 07/Jun/25
$$\mathrm{sir}\:\mathrm{mrW}\:\mathrm{and}\:\mathrm{sir}\:\mathrm{Aleks}\:\mathrm{solved}\:\mathrm{a}\:\mathrm{problem}\:\mathrm{too} \\ $$$$\mathrm{sometimes}\:\mathrm{ago}.\:\mathrm{Do}\:\mathrm{you}\:\mathrm{remember}\:\mathrm{sir}. \\ $$$$\mathrm{It}\:\mathrm{leaads}\:\mathrm{to}\:\mathrm{integration}\:\mathrm{or}\:\mathrm{so}. \\ $$$$\mathrm{Do}\:\mathrm{you}\:\mathrm{still}\:\mathrm{remember}\:\mathrm{the}\:\mathrm{Question}\:\mathrm{number}\:\mathrm{sir}? \\ $$
Commented by mr W last updated on 07/Jun/25
$${you}\:{seem}\:{not}\:{to}\:{have}\:{known}\:{that} \\ $$$${ajfour}\:{sir}\:{has}\:{solved}\:{this}\:{famous} \\ $$$${problem}.\:{this}\:{is}\:{strange}!\:{actually}\:{i} \\ $$$${can}\:{remember}\:{that}\:{you}\:{also}\:{have} \\ $$$${congratulated}\:{them}\:{at}\:{that}\:{time}. \\ $$
Commented by Tawa11 last updated on 07/Jun/25
$$\mathrm{I}\:\mathrm{think}\:\mathrm{I}\:\mathrm{remember}\:\mathrm{that}\:\mathrm{time}\:\mathrm{sir}. \\ $$
Commented by Tawa11 last updated on 07/Jun/25
$$\mathrm{Do}\:\mathrm{you}\:\mathrm{remember}\:\mathrm{your}\:\mathrm{own}\:\mathrm{solution} \\ $$$$\mathrm{and}\:\mathrm{sir}\:\mathrm{aleks}\:\mathrm{question}\:\mathrm{number}\:\mathrm{sir}? \\ $$
Commented by mr W last updated on 07/Jun/25
$${i}\:{solved}\:{a}\:{lot}\:{of}\:{questions}\:{in}\:{the} \\ $$$${forum}.\:{i}\:{don}'{t}\:{know}\:{which}\:{question} \\ $$$${you}\:{mean}. \\ $$
Commented by Rasheed.Sindhi last updated on 07/Jun/25
$$'{Frix}'\:{is}\:{new}\:{identity}\:{of}\:{mjs}\:{sir}, \\ $$$${I}\:{think}. \\ $$$${Sir}\:{mrW}\:,\:{Please}\:{write}\:{a}\:{questionID} \\ $$$${in}\:{which}\:{sir}\:{ajfour}\:{and}\:{sir}\:{mjs} \\ $$$${received}\:{congratulation}\:{on}\:{solving} \\ $$$${the}\:{above}\:{question},\:{if}\:{you}\:{know}. \\ $$
Commented by ajfour last updated on 07/Jun/25
$${They}\:{gave}\:{us}\:{not}\:{one}\:{penny}.\:{invited} \\ $$$${us}\:{on}\:{slack}.{com}\:{and}\:{later}\:{said}\: \\ $$$${project}\:{was}\:{with}\:{nasa}\:{and}\:{they}\:{closed}\:{it} \\ $$$${and}\:{that}\:{merit}\:{of}\:{the}\:{solution} \\ $$$$\:{needed}\:{to}\:{be}\:{evaluated}.\: \\ $$$${Lycon}\:{Trix} \\ $$$${Alphaprime} \\ $$$${Siddharth} \\ $$$${above}\:{named}\:{guys}\:{belonged}\:{to}\:{those} \\ $$$${seeking}\:{the}\:{solution}.\: \\ $$
Commented by Tawa11 last updated on 07/Jun/25
$$\mathrm{This}\:\mathrm{life}\:\mathrm{is}\:\mathrm{full}\:\mathrm{of}\:\mathrm{cheating}. \\ $$
Commented by ajfour last updated on 07/Jun/25
Commented by ajfour last updated on 07/Jun/25
$${Q}.\:\mathrm{60723} \\ $$
Commented by Tawa11 last updated on 07/Jun/25
$$\mathrm{Sir}\:\mathrm{ajfour}. \\ $$$$\mathrm{They}\:\mathrm{stole}\:\mathrm{your}\:\mathrm{result}. \\ $$$$\mathrm{Probably}\:\mathrm{they}\:\mathrm{have}\:\mathrm{claim}\:\mathrm{the}\:\mathrm{money}. \\ $$$$\mathrm{Cheater}. \\ $$
Commented by Tawa11 last updated on 07/Jun/25
$$\mathrm{Ajfour},\:\mathrm{have}\:\mathrm{you}\:\mathrm{sent}\:\mathrm{the}\:\mathrm{answer}? \\ $$$$\mathrm{Can}\:\mathrm{I}\:\mathrm{email}\:\mathrm{you}\:\mathrm{to}\:\mathrm{share}\:\mathrm{me}. \\ $$$$\mathrm{But}\:\mathrm{if}\:\mathrm{you}\:\mathrm{have}\:\mathrm{not}\:\mathrm{share}\:\mathrm{the}\:\mathrm{result}. \\ $$$$\mathrm{No}\:\mathrm{need}\:\mathrm{to}\:\mathrm{share}\:\mathrm{with}\:\mathrm{me}. \\ $$
Commented by Frix last updated on 08/Jun/25
$$\mathrm{I}'\mathrm{m}\:\mathrm{only}\:\mathrm{myself}.\:\mathrm{Sorry}\:\mathrm{Sirs}\:\mathrm{and}\:\mathrm{Ladies}. \\ $$
Commented by mr W last updated on 08/Jun/25
$$@{Frix}\:{sir}:\: \\ $$$${thanks}\:{for}\:{the}\:{clarification}! \\ $$
Commented by mr W last updated on 08/Jun/25
$$@{Rasheed}\:{sir}: \\ $$$${that}'{s}\:{the}\:{Q}\mathrm{61490}\:{where}\:{MJS}\:{sir} \\ $$$${also}\:{shared}\:{their}\:{complete} \\ $$$${solution}. \\ $$
Commented by Ghisom last updated on 08/Jun/25
$$\mathrm{great}! \\ $$
Commented by Rasheed.Sindhi last updated on 09/Jun/25
$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{mr}}\mathbb{W}! \\ $$
Commented by Tawa11 last updated on 12/Jun/25
Commented by Tawa11 last updated on 12/Jun/25
$$\mathrm{Ajfour}\:\mathrm{sir}. \\ $$$$\mathrm{Have}\:\mathrm{you}\:\mathrm{completed}\:\mathrm{this}\:\mathrm{work}\:\mathrm{sir}? \\ $$
Answered by Ghisom last updated on 08/Jun/25
![looking at the given solution I think I know how Aifour & MJS got there (√(a−(√(a+x))))+(√(a+(√(a−x))))=2x staying in R ⇒ (√(a+x))≥(√(a−x)) let (√(a+x))=α+β∧(√(a−x))=α−β ⇔ α=((√(a+x))/2)+((√(a−x))/2)∧β=((√(a+x))/2)−((√(a−x))/2) ⇒ x=2αβ∧a=α^2 +β^2 inserting we get (√(a−α−β))+(√(a+α−β))=4αβ squaring & transforming (√(a−α−β))(√(a+α−β))=8α^2 β^2 +β−a squaring & transforming α^2 (64α^2 β^4 +16β^3 −16aβ^2 +1)=0 α≠0 [or else x=a=0] 64α^2 β^4 +16β^3 −16aβ^2 +1=0 using a=α^2 +β^2 ⇔ α^2 =a−β^2 64β^6 −64aβ^4 −16β^3 +16aβ^2 −1=0 64β^6 −16β^3 −1=64aβ^4 +16aβ^2 let β=(γ/2) γ^6 −2γ^3 −1=4a(γ^4 −γ^2 ) β>0 ⇒ γ>0 γ^3 −(1/γ^3 )−2=4a(γ−(1/γ)) γ^3 −(1/γ^3 )=(γ−(1/γ))^3 +3(γ−(1/γ)) let γ−(1/γ)=δ δ^3 +3δ−2=4aδ δ^3 +(3−4a)δ−2=0 this can be solved using the Trigonometric Solution Method. but it gives 3 solutions, how to know which is the right one? also for a≤≈1.5 there′s no solution to the original equation although we seem to get one... [my δ is the same as Aifour′s & MJS′s r]](https://www.tinkutara.com/question/Q221636.png)
$$\mathrm{looking}\:\mathrm{at}\:\mathrm{the}\:\mathrm{given}\:\mathrm{solution}\:\mathrm{I}\:\mathrm{think}\:\mathrm{I}\:\mathrm{know} \\ $$$$\mathrm{how}\:\mathrm{Aifour}\:\&\:\mathrm{MJS}\:\mathrm{got}\:\mathrm{there} \\ $$$$ \\ $$$$\sqrt{{a}−\sqrt{{a}+{x}}}+\sqrt{{a}+\sqrt{{a}−{x}}}=\mathrm{2}{x} \\ $$$$\mathrm{staying}\:\mathrm{in}\:\mathbb{R}\:\Rightarrow\:\sqrt{{a}+{x}}\geqslant\sqrt{{a}−{x}} \\ $$$$\mathrm{let}\:\sqrt{{a}+{x}}=\alpha+\beta\wedge\sqrt{{a}−{x}}=\alpha−\beta \\ $$$$\Leftrightarrow \\ $$$$\alpha=\frac{\sqrt{{a}+{x}}}{\mathrm{2}}+\frac{\sqrt{{a}−{x}}}{\mathrm{2}}\wedge\beta=\frac{\sqrt{{a}+{x}}}{\mathrm{2}}−\frac{\sqrt{{a}−{x}}}{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${x}=\mathrm{2}\alpha\beta\wedge{a}=\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{inserting}\:\mathrm{we}\:\mathrm{get} \\ $$$$\sqrt{{a}−\alpha−\beta}+\sqrt{{a}+\alpha−\beta}=\mathrm{4}\alpha\beta \\ $$$$\mathrm{squaring}\:\&\:\mathrm{transforming} \\ $$$$\sqrt{{a}−\alpha−\beta}\sqrt{{a}+\alpha−\beta}=\mathrm{8}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} +\beta−{a} \\ $$$$\mathrm{squaring}\:\&\:\mathrm{transforming} \\ $$$$\alpha^{\mathrm{2}} \left(\mathrm{64}\alpha^{\mathrm{2}} \beta^{\mathrm{4}} +\mathrm{16}\beta^{\mathrm{3}} −\mathrm{16}{a}\beta^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$\alpha\neq\mathrm{0}\:\left[\mathrm{or}\:\mathrm{else}\:{x}={a}=\mathrm{0}\right] \\ $$$$\mathrm{64}\alpha^{\mathrm{2}} \beta^{\mathrm{4}} +\mathrm{16}\beta^{\mathrm{3}} −\mathrm{16}{a}\beta^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{using}\:{a}=\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \:\Leftrightarrow\:\alpha^{\mathrm{2}} ={a}−\beta^{\mathrm{2}} \\ $$$$\mathrm{64}\beta^{\mathrm{6}} −\mathrm{64}{a}\beta^{\mathrm{4}} −\mathrm{16}\beta^{\mathrm{3}} +\mathrm{16}{a}\beta^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{64}\beta^{\mathrm{6}} −\mathrm{16}\beta^{\mathrm{3}} −\mathrm{1}=\mathrm{64}{a}\beta^{\mathrm{4}} +\mathrm{16}{a}\beta^{\mathrm{2}} \\ $$$$\mathrm{let}\:\beta=\frac{\gamma}{\mathrm{2}} \\ $$$$\gamma^{\mathrm{6}} −\mathrm{2}\gamma^{\mathrm{3}} −\mathrm{1}=\mathrm{4}{a}\left(\gamma^{\mathrm{4}} −\gamma^{\mathrm{2}} \right) \\ $$$$\beta>\mathrm{0}\:\Rightarrow\:\gamma>\mathrm{0} \\ $$$$\gamma^{\mathrm{3}} −\frac{\mathrm{1}}{\gamma^{\mathrm{3}} }−\mathrm{2}=\mathrm{4}{a}\left(\gamma−\frac{\mathrm{1}}{\gamma}\right) \\ $$$$\gamma^{\mathrm{3}} −\frac{\mathrm{1}}{\gamma^{\mathrm{3}} }=\left(\gamma−\frac{\mathrm{1}}{\gamma}\right)^{\mathrm{3}} +\mathrm{3}\left(\gamma−\frac{\mathrm{1}}{\gamma}\right) \\ $$$$\mathrm{let}\:\gamma−\frac{\mathrm{1}}{\gamma}=\delta \\ $$$$\delta^{\mathrm{3}} +\mathrm{3}\delta−\mathrm{2}=\mathrm{4}{a}\delta \\ $$$$\delta^{\mathrm{3}} +\left(\mathrm{3}−\mathrm{4}{a}\right)\delta−\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{this}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{using}\:\mathrm{the}\:\mathrm{Trigonometric} \\ $$$$\mathrm{Solution}\:\mathrm{Method}.\:\mathrm{but}\:\mathrm{it}\:\mathrm{gives}\:\mathrm{3}\:\mathrm{solutions}, \\ $$$$\mathrm{how}\:\mathrm{to}\:\mathrm{know}\:\mathrm{which}\:\mathrm{is}\:\mathrm{the}\:\mathrm{right}\:\mathrm{one}? \\ $$$$\mathrm{also}\:\mathrm{for}\:{a}\leqslant\approx\mathrm{1}.\mathrm{5}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{original}\:\mathrm{equation}\:\mathrm{although}\:\mathrm{we}\:\mathrm{seem}\:\mathrm{to}\:\mathrm{get} \\ $$$$\mathrm{one}… \\ $$$$ \\ $$$$\left[\mathrm{my}\:\delta\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\mathrm{Aifour}'\mathrm{s}\:\&\:\mathrm{MJS}'\mathrm{s}\:{r}\right] \\ $$
Answered by Jubr last updated on 12/Jun/25
Commented by Jubr last updated on 12/Jun/25
$${Please}\:{is}\:{it}\:{correct}? \\ $$
Commented by Frix last updated on 12/Jun/25
$$\theta\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{parameter}\:\mathrm{but}\:\mathrm{a}\:\mathrm{function}\:\mathrm{of}\:{x} \\ $$$$\Rightarrow\:\mathrm{It}'\mathrm{s}\:\mathrm{not}\:\mathrm{solved}. \\ $$
Commented by Tawa11 last updated on 12/Jun/25
Commented by Tawa11 last updated on 12/Jun/25
$$\mathrm{Ajfour}\:\mathrm{sir}. \\ $$$$\mathrm{Did}\:\mathrm{you}\:\mathrm{complete}\:\mathrm{this}\:\mathrm{work}? \\ $$
Commented by Jubr last updated on 13/Jun/25
Commented by Jubr last updated on 13/Jun/25
Commented by Jubr last updated on 13/Jun/25
Commented by Jubr last updated on 13/Jun/25
$${Please}\:{sir}.\:{What}\:{of}\:{this}\:{solution}. \\ $$$${Is}\:{this}\:{correct}?? \\ $$$${Thanks}\:{for}\:{previous}\:{feedback}. \\ $$
Commented by Ghisom last updated on 13/Jun/25
$$\mathrm{check}\:\mathrm{2}\:\mathrm{things}: \\ $$$$\mathrm{1}.\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{a}<\mathrm{1}.\mathrm{509830340885}… \\ $$$$\:\:\:\:\:{a}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{of} \\ $$$$\:\:\:\:\:{t}^{\mathrm{6}} −\mathrm{3}{t}^{\mathrm{5}} +\mathrm{3}{t}^{\mathrm{4}} −\frac{\mathrm{3}}{\mathrm{2}}{t}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{0} \\ $$$$\mathrm{2}.\:\mathrm{there}'\mathrm{s}\:\mathrm{an}\:\mathrm{exact}\:\mathrm{solution}\:\mathrm{for}\:{a}=\mathrm{2}: \\ $$$$\:\:\:\:\:{x}=\sqrt{\frac{\mathrm{9}}{\mathrm{8}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{13}+\mathrm{8}\sqrt{\mathrm{2}}}}{\mathrm{8}}}\approx\mathrm{1}.\mathrm{10260832651} \\ $$
Commented by Jubr last updated on 13/Jun/25
$${Thanks}\:{sir}. \\ $$
Commented by Tawa11 last updated on 13/Jun/25
$$\mathrm{Sir},\:\mathrm{please},\:\mathrm{how}\:\mathrm{can}\:\mathrm{one}\:\mathrm{think}\:\mathrm{of}\:\mathrm{the}\:\mathrm{a}_{\mathrm{0}} . \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{setup}\:\mathrm{to}\:\mathrm{get}\:\:\:\:\mathrm{t}^{\mathrm{6}} \:−\:\mathrm{3t}^{\mathrm{5}} \:+\:\mathrm{3t}^{\mathrm{4}} \:−\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{t}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{t}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{\mathrm{8}}\:=\:\:\mathrm{0} \\ $$
Commented by Ghisom last updated on 14/Jun/25
![you can use software to draw both f(x)=(√(a−(√(a+x))))+(√(a+(√(a−x)))) g(x)=2x for some values of a if a<≈(3/2) the graphs don′t intersect we must find for which values of a, x f(x) ∈R obviously x≥0 a+x≥0∧a−x≥0 ⇒ −a≤x≤a but if a<0 ⇒ a−(√(a+x))<0 ⇒ a≥0 ⇒ a≥0∧0≤x≤a ⇒ a+(√(a−x))≥0 also needed: a−(√(a+x))≥0 a≥(√(a+x)) both sides ≥0 ⇒ we can square a^2 ≥a+x a^2 −a≥x x≤a(a−1) now we have a≥0∧x≤a∧x≤a(a−1) ⇔ a≥0∧0≤x≤min (a, a(a−1)) if x=a (√(a−(√(a+a))))+(√a)=2a ⇒ a=0 as only solution ⇒ x=a=0 if x=a(a−1) a^(1/4) (√((√a)+(√(2−a))))=2a(a−1) a=0 ⇒ x=a=0 (√((√a)+(√(2−a))))=2a^(3/4) (a−1) (√a)+(√(2−a))=2a^(3/2) (a−1) (√(2−a))=2a(√a)(a−1)−(√a) (√(2−a))=(√a)(4a^3 −8a^2 +4a−1) 2−a=a(4a^3 −8a^2 +4a−1)^2 a^7 −4a^6 +6a^5 −(9/2)a^4 +2a^3 −(1/2)a^2 +(1/8)a−(1/8)=0 (a−1)(a^6 −3a^5 +3a^4 −(3/2)a^3 +(1/2)a^2 +(1/8))=0 a=1 ⇒ x=0 which is a false solution [we squared!] a^6 −3a^5 +3a^4 −(3/2)a^3 +(1/2)a^2 +(1/8)=0 ⇒ a≈1.19281015388... which again is false a=a_0 ≈1.50983034088... ⇒ x≈.769757317372... it turns out there′s no solution for x if a<a_0](https://www.tinkutara.com/question/Q221967.png)
$$\mathrm{you}\:\mathrm{can}\:\mathrm{use}\:\mathrm{software}\:\mathrm{to}\:\mathrm{draw}\:\mathrm{both} \\ $$$${f}\left({x}\right)=\sqrt{{a}−\sqrt{{a}+{x}}}+\sqrt{{a}+\sqrt{{a}−{x}}} \\ $$$${g}\left({x}\right)=\mathrm{2}{x} \\ $$$$\mathrm{for}\:\mathrm{some}\:\mathrm{values}\:\mathrm{of}\:{a} \\ $$$$\mathrm{if}\:{a}<\approx\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{the}\:\mathrm{graphs}\:\mathrm{don}'\mathrm{t}\:\mathrm{intersect} \\ $$$$\mathrm{we}\:\mathrm{must}\:\mathrm{find}\:\mathrm{for}\:\mathrm{which}\:\mathrm{values}\:\mathrm{of}\:{a},\:{x} \\ $$$${f}\left({x}\right)\:\in\mathbb{R} \\ $$$$\mathrm{obviously} \\ $$$${x}\geqslant\mathrm{0} \\ $$$${a}+{x}\geqslant\mathrm{0}\wedge{a}−{x}\geqslant\mathrm{0}\:\Rightarrow\:−{a}\leqslant{x}\leqslant{a} \\ $$$$\mathrm{but}\:\mathrm{if}\:{a}<\mathrm{0}\:\Rightarrow\:{a}−\sqrt{{a}+{x}}<\mathrm{0}\:\Rightarrow\:{a}\geqslant\mathrm{0} \\ $$$$\Rightarrow \\ $$$${a}\geqslant\mathrm{0}\wedge\mathrm{0}\leqslant{x}\leqslant{a} \\ $$$$\Rightarrow\:{a}+\sqrt{{a}−{x}}\geqslant\mathrm{0} \\ $$$$\mathrm{also}\:\mathrm{needed}:\:{a}−\sqrt{{a}+{x}}\geqslant\mathrm{0} \\ $$$${a}\geqslant\sqrt{{a}+{x}}\:\:\:\:\:\mathrm{both}\:\mathrm{sides}\:\geqslant\mathrm{0}\:\Rightarrow\:\mathrm{we}\:\mathrm{can}\:\mathrm{square} \\ $$$${a}^{\mathrm{2}} \geqslant{a}+{x} \\ $$$${a}^{\mathrm{2}} −{a}\geqslant{x} \\ $$$${x}\leqslant{a}\left({a}−\mathrm{1}\right) \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{have} \\ $$$${a}\geqslant\mathrm{0}\wedge{x}\leqslant{a}\wedge{x}\leqslant{a}\left({a}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\Leftrightarrow\:{a}\geqslant\mathrm{0}\wedge\mathrm{0}\leqslant{x}\leqslant\mathrm{min}\:\left({a},\:{a}\left({a}−\mathrm{1}\right)\right) \\ $$$$\mathrm{if}\:{x}={a} \\ $$$$\sqrt{{a}−\sqrt{{a}+{a}}}+\sqrt{{a}}=\mathrm{2}{a} \\ $$$$\Rightarrow\:{a}=\mathrm{0}\:\mathrm{as}\:\mathrm{only}\:\mathrm{solution}\:\Rightarrow\:{x}={a}=\mathrm{0} \\ $$$$\mathrm{if}\:{x}={a}\left({a}−\mathrm{1}\right) \\ $$$${a}^{\mathrm{1}/\mathrm{4}} \sqrt{\sqrt{{a}}+\sqrt{\mathrm{2}−{a}}}=\mathrm{2}{a}\left({a}−\mathrm{1}\right) \\ $$$${a}=\mathrm{0}\:\Rightarrow\:{x}={a}=\mathrm{0} \\ $$$$\sqrt{\sqrt{{a}}+\sqrt{\mathrm{2}−{a}}}=\mathrm{2}{a}^{\mathrm{3}/\mathrm{4}} \left({a}−\mathrm{1}\right) \\ $$$$\sqrt{{a}}+\sqrt{\mathrm{2}−{a}}=\mathrm{2}{a}^{\mathrm{3}/\mathrm{2}} \left({a}−\mathrm{1}\right) \\ $$$$\sqrt{\mathrm{2}−{a}}=\mathrm{2}{a}\sqrt{{a}}\left({a}−\mathrm{1}\right)−\sqrt{{a}} \\ $$$$\sqrt{\mathrm{2}−{a}}=\sqrt{{a}}\left(\mathrm{4}{a}^{\mathrm{3}} −\mathrm{8}{a}^{\mathrm{2}} +\mathrm{4}{a}−\mathrm{1}\right) \\ $$$$\mathrm{2}−{a}={a}\left(\mathrm{4}{a}^{\mathrm{3}} −\mathrm{8}{a}^{\mathrm{2}} +\mathrm{4}{a}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$${a}^{\mathrm{7}} −\mathrm{4}{a}^{\mathrm{6}} +\mathrm{6}{a}^{\mathrm{5}} −\frac{\mathrm{9}}{\mathrm{2}}{a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{8}}{a}−\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{0} \\ $$$$\left({a}−\mathrm{1}\right)\left({a}^{\mathrm{6}} −\mathrm{3}{a}^{\mathrm{5}} +\mathrm{3}{a}^{\mathrm{4}} −\frac{\mathrm{3}}{\mathrm{2}}{a}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{8}}\right)=\mathrm{0} \\ $$$${a}=\mathrm{1}\:\Rightarrow\:{x}=\mathrm{0}\:\mathrm{which}\:\mathrm{is}\:\mathrm{a}\:\mathrm{false}\:\mathrm{solution}\:\left[\mathrm{we}\right. \\ $$$$\left.\mathrm{squared}!\right] \\ $$$${a}^{\mathrm{6}} −\mathrm{3}{a}^{\mathrm{5}} +\mathrm{3}{a}^{\mathrm{4}} −\frac{\mathrm{3}}{\mathrm{2}}{a}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${a}\approx\mathrm{1}.\mathrm{19281015388}…\:\mathrm{which}\:\mathrm{again}\:\mathrm{is}\:\mathrm{false} \\ $$$$ \\ $$$${a}={a}_{\mathrm{0}} \approx\mathrm{1}.\mathrm{50983034088}…\:\Rightarrow\:{x}\approx.\mathrm{769757317372}… \\ $$$$\mathrm{it}\:\mathrm{turns}\:\mathrm{out}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{x}\:\mathrm{if}\:{a}<{a}_{\mathrm{0}} \\ $$
Commented by Tawa11 last updated on 14/Jun/25
$$\mathrm{Wow},\:\mathrm{I}\:\mathrm{understand}\:\mathrm{now}\:\mathrm{sir}. \\ $$