Question Number 211255 by ajfour last updated on 02/Sep/24
$$\sqrt{{a}+\sqrt{{b}−{x}}+\sqrt{{b}−\sqrt{{a}+{x}}}}=\mathrm{2}{x} \\ $$$${solve}\:{for}\:{x}.\:\:\:\: \\ $$
Commented by Ghisom last updated on 03/Sep/24
![(√(a+(√(b−x))+(√(b−(√(a+x))))))=2x x=(b−(4x^2 −(√(b−x))−a)^2 )^2 −a it′s already of 8^(th) degree and we still have (√(b−x)) ⇒ it′s of 16^(th) degree. we can′t solve for x nor for a. but we can for b: b=((256x^8 −256ax^6 +32x^5 +96a^2 x^4 −16ax^3 −(16a^3 −1)x^2 +(2a^2 +1)x+a(a^3 +1))/(4(4x^2 −a)^2 ))+(((16x^4 −8ax^2 −x+a^2 )(√(a+x)))/(2(4x^2 −a)^2 )) [might not always be true because of 3 times squaring]](https://www.tinkutara.com/question/Q211271.png)
$$\sqrt{{a}+\sqrt{{b}−{x}}+\sqrt{{b}−\sqrt{{a}+{x}}}}=\mathrm{2}{x} \\ $$$${x}=\left({b}−\left(\mathrm{4}{x}^{\mathrm{2}} −\sqrt{{b}−{x}}−{a}\right)^{\mathrm{2}} \right)^{\mathrm{2}} −{a} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{already}\:\mathrm{of}\:\mathrm{8}^{\mathrm{th}} \:\mathrm{degree}\:\mathrm{and}\:\mathrm{we}\:\mathrm{still}\:\mathrm{have} \\ $$$$\sqrt{{b}−{x}}\:\Rightarrow\:\mathrm{it}'\mathrm{s}\:\mathrm{of}\:\mathrm{16}^{\mathrm{th}} \:\mathrm{degree}.\:\mathrm{we}\:\mathrm{can}'\mathrm{t}\:\mathrm{solve} \\ $$$$\mathrm{for}\:{x}\:\mathrm{nor}\:\mathrm{for}\:{a}.\:\mathrm{but}\:\mathrm{we}\:\mathrm{can}\:\mathrm{for}\:{b}: \\ $$$${b}=\frac{\mathrm{256}{x}^{\mathrm{8}} −\mathrm{256}{ax}^{\mathrm{6}} +\mathrm{32}{x}^{\mathrm{5}} +\mathrm{96}{a}^{\mathrm{2}} {x}^{\mathrm{4}} −\mathrm{16}{ax}^{\mathrm{3}} −\left(\mathrm{16}{a}^{\mathrm{3}} −\mathrm{1}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1}\right){x}+{a}\left({a}^{\mathrm{3}} +\mathrm{1}\right)}{\mathrm{4}\left(\mathrm{4}{x}^{\mathrm{2}} −{a}\right)^{\mathrm{2}} }+\frac{\left(\mathrm{16}{x}^{\mathrm{4}} −\mathrm{8}{ax}^{\mathrm{2}} −{x}+{a}^{\mathrm{2}} \right)\sqrt{{a}+{x}}}{\mathrm{2}\left(\mathrm{4}{x}^{\mathrm{2}} −{a}\right)^{\mathrm{2}} } \\ $$$$\left[\mathrm{might}\:\mathrm{not}\:\mathrm{always}\:\mathrm{be}\:\mathrm{true}\:\mathrm{because}\:\mathrm{of}\:\mathrm{3}\right. \\ $$$$\left.\mathrm{times}\:\mathrm{squaring}\right] \\ $$
Commented by ajfour last updated on 06/Sep/24
$$\sqrt{{a}+\sqrt{{b}−{x}}}+\sqrt{{b}−\sqrt{{a}+{x}}}=\mathrm{2}{x} \\ $$$${i}\:{think}\:{i}\:{was}\:{interested}\:{in}\:{this}!\:{rather} \\ $$
Commented by ajfour last updated on 06/Sep/24
![let a+x=p^2 , b−x=q^2 p^2 +q^2 =a+b 2x=(p^2 −q^2 )−(a−b) (√(a+q))+(√(b−p))=p^2 −q^2 −(a−b) squaring a+q={2(p^2 −a)−(√(b−p))}^2 a+b−p^2 =[{2(p^2 −a)−(√(b−p))}^2 −a]^2 .....](https://www.tinkutara.com/question/Q211361.png)
$${let}\:\:{a}+{x}={p}^{\mathrm{2}} \:\:,\:\:{b}−{x}={q}^{\mathrm{2}} \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} ={a}+{b} \\ $$$$\mathrm{2}{x}=\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)−\left({a}−{b}\right) \\ $$$$\sqrt{{a}+{q}}+\sqrt{{b}−{p}}={p}^{\mathrm{2}} −{q}^{\mathrm{2}} −\left({a}−{b}\right) \\ $$$${squaring} \\ $$$${a}+{q}=\left\{\mathrm{2}\left({p}^{\mathrm{2}} −{a}\right)−\sqrt{{b}−{p}}\right\}^{\mathrm{2}} \\ $$$${a}+{b}−{p}^{\mathrm{2}} =\left[\left\{\mathrm{2}\left({p}^{\mathrm{2}} −{a}\right)−\sqrt{{b}−{p}}\right\}^{\mathrm{2}} −{a}\right]^{\mathrm{2}} \\ $$$$….. \\ $$
Answered by a.lgnaoui last updated on 03/Sep/24
![4x^2 −a = (√(b−x)) +(√(b−(√(a+x)))) (4x^2 −a)−(√(b−x)) =(√(b−(√(a+x)))) (4x^2 −a)^2 −x−2(4x^2 −a)(√(b−x)) = −(√(a+x)) ((4x^2 −a)^2 −x=2(4x^2 −a)(√(b−x)) −(√(a+x)) 4x^2 −a)^2 +(√(a+x)) =2(4x^2 −a)(√(b−x)) ((4x^2 −a)/2)+((√(a+x))/(2(4x^2 −a)))=(√(b−x)) (((4x^2 −a)^2 )/4)+((a+x)/(4(4x^2 −a)^2 ))+(((4x^2 −a)(√(a+x)))/2)=b−x (√(a+x)) =(2/((4x^2 −a)))[b−x−((a+x)/(4(4x^2 −a)^2 ))−(((4x^2 −a)^2 )/4)] (√(a+x)) =(2/(4x^2 −a))[b−x−((([(a+x)+(4x^2 −a)^4 )/(4(4x^2 −a)^2 ))) =[((2(b−x)(4x^2 −a)−(a+x)−(4x^2 −a)^4 )/((4x^2 −a)^2 ))] =(([(4x^2 −a)[(b−x)−(4x^2 −a)^3 ])/((4x^2 −a)^2 )) a+x=(((4x^2 −a)^2 (b−x−(4x^2 −a)^3 )^2 )/((4x^2 −a)^4 )) (4x^2 −a)^4 (a+x)= (4x^2 −a)^2 [(b−x)^2 +(4x^2 −a)^6 −2(b−x)(4x^2 −a)^3 ] (4x^2 −a)^2 (a+x) = (b−x)^2 +(4x^2 −a)^6 −2(b−x)(4x^2 −a)^3 ] (4x^2 −a)^2 (a+x)−(4x^2 −a)^6 +2(b−x)(4x^2 −a)^3 =(b−x)^2 (4x^2 −a)^2 [a+x−(4x^2 −a)^4 +2(b−x)(4x^2 −a)] =(b−x)^2 [a+x+8bx^2 −2ab−8x^3 +2ax −(4x^2 −a)^4 =(b−x)^2 (b−x)^2 +(4x^2 −a)^4 = [(b−x+(4x^2 −a)^2 ]^2 −2(b−x)(4x^2 −a)^2 soit −8x^3 +8bx^2 +(2a+1)x+a(1−2b)= b^2 +x^2 −2bx+(16x^4 +a^2 −8ax^2 )^2 +2(b−x)(4x^2 −a)^2 =x^2 −2bx +b^2 +[8x^2 (a+2x^2 )+a^2 ] +64x^4 (a+2x^2 )^2 +a^4 +16a^2 (a+2x^2 ) +2b−x)(16x^4 +a^2 −8ax^2 ) ⇒ (256+a)x^6 −16x^5 +(64a^2 +32b+8a)x^4 +8(a+1)x^3 (32a^2 −16ab−8)x^2 −(a^2 +2a+1)x+16a^3 +a^4 −a(1−2b)=0 256(1+a)x^6 −16x^5 +8(8a^2 +4b+a)x^4 +8(a+1)x^3 +8(4a^2 −2ab−1)x^2 −(a+1)^2 x +a^4 +16a^3 −a+2ab=0 (256x^6 −16x^5 +8x^3 −8x^2 −x)+ 256ax^6 +8(8a^2 +4b+a)x^4 +8ax^3 +16(2a^2 −ab)x^2 −(a^2 +2a)x+ a^4 +16a^3 −a+2ab=0 { ((256x^5 −16x^4 +8x^2 +8x−1=0)),((8[32ax^6 +(8a^2 +4b+a)x^4 +ax^3 +)) :} a(4a−b)x^2 −a(a+2)x]+ a(a^3 +26a^2 −1+2b)=0 { ((x=0,112166986... and imsg(x) (1))),((8ax(32x^5 +x^2 +(4a−b)x−(a+2)=0(2))) :} •3 a^3 +16a^2 +2b−1=0 1• x= 2et 3• { ((32x^5 +x^2 +(4a−b)x=a+2)),(( a^3 +16a^2 +2b−1 =0)) :} a^2 (a+16)=1−2b b=((1−a^2 (a+16))/2) 4a−b=((a^2 (a+16)+8a−1)/2) (2)⇒ 32x^5 +x^2 +((a^2 (a+16)+8a−1)/2)x=a+2 we change x by x_0 =0,1121167. ⇒we bave equation with a as unconnu then we found a and b (exept Ereur in the Calcul)](https://www.tinkutara.com/question/Q211268.png)
$$\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\sqrt{\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}}\:+\sqrt{\boldsymbol{\mathrm{b}}−\sqrt{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}}} \\ $$$$\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)−\sqrt{\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}}\:=\sqrt{\boldsymbol{\mathrm{b}}−\sqrt{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}}} \\ $$$$\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} −\boldsymbol{\mathrm{x}}−\mathrm{2}\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)\sqrt{\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}}\:= \\ $$$$−\sqrt{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}} \\ $$$$ \\ $$$$\left(\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} −\boldsymbol{\mathrm{x}}=\mathrm{2}\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)\sqrt{\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}}\:−\sqrt{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}}\right. \\ $$$$ \\ $$$$\left.\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} +\sqrt{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}}\:=\mathrm{2}\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)\sqrt{\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}} \\ $$$$\frac{\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}}{\mathrm{2}}+\frac{\sqrt{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}}}{\mathrm{2}\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)}=\sqrt{\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}} \\ $$$$\frac{\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}}{\mathrm{4}\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} }+\frac{\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)\sqrt{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}}}{\mathrm{2}}=\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}} \\ $$$$ \\ $$$$\sqrt{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}}\:=\frac{\mathrm{2}}{\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)}\left[\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}−\frac{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}}{\mathrm{4}\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} }−\frac{\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} }{\mathrm{4}}\right] \\ $$$$ \\ $$$$\sqrt{\mathrm{a}+\boldsymbol{\mathrm{x}}}\:=\frac{\mathrm{2}}{\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}}\left[\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}−\left(\frac{\left[\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}\right)+\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{4}} \right.}{\mathrm{4}\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} }\right)\right. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:=\left[\frac{\mathrm{2}\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}\right)\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)−\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}\right)−\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{4}} }{\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} }\right] \\ $$$$=\frac{\left[\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)\left[\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}\right)−\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{3}} \right]\right.}{\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}=\frac{\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} \left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}−\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{3}} \right)^{\mathrm{2}} }{\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{4}} } \\ $$$$ \\ $$$$ \\ $$$$\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{a}\right)^{\mathrm{4}} \left(\mathrm{a}+\boldsymbol{\mathrm{x}}\right)= \\ $$$$\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} \left[\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}\right)^{\mathrm{2}} +\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{6}} −\mathrm{2}\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}\right)\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{3}} \right] \\ $$$$ \\ $$$$ \\ $$$$\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} \left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}\right)\:\:= \\ $$$$\left.\:\:\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}\right)^{\mathrm{2}} +\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{6}} −\mathrm{2}\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}\right)\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{3}} \right] \\ $$$$ \\ $$$$\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} \left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}\right)−\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{6}} +\mathrm{2}\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}\right)\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{3}} \\ $$$$=\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} \left[\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}−\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{4}} +\mathrm{2}\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}\right)\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)\right] \\ $$$$ \\ $$$$=\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}\right)^{\mathrm{2}} \\ $$$$\left[\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}+\mathrm{8}\boldsymbol{\mathrm{bx}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{ab}}−\mathrm{8}\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{2}\boldsymbol{\mathrm{ax}}\right. \\ $$$$−\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{4}} =\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}\right)^{\mathrm{2}} +\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{a}\right)^{\mathrm{4}} = \\ $$$$ \\ $$$$\left[\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}+\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} \right]^{\mathrm{2}} −\mathrm{2}\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}\right)\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} \right. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{soit} \\ $$$$\:\:−\mathrm{8}\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{8}\boldsymbol{\mathrm{bx}}^{\mathrm{2}} +\left(\mathrm{2}\boldsymbol{\mathrm{a}}+\mathrm{1}\right)\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{a}}\left(\mathrm{1}−\mathrm{2}\boldsymbol{\mathrm{b}}\right)= \\ $$$$\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{bx}}+\left(\mathrm{16}\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\mathrm{8}\boldsymbol{\mathrm{ax}}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$+\mathrm{2}\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}\right)\left(\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$=\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{bx}}\:+\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\left[\mathrm{8}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{a}}+\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)+\boldsymbol{\mathrm{a}}^{\mathrm{2}} \right] \\ $$$$\:\:\:+\mathrm{64}\boldsymbol{\mathrm{x}}^{\mathrm{4}} \left(\boldsymbol{\mathrm{a}}+\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{4}} +\mathrm{16}\boldsymbol{\mathrm{a}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{a}}+\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right) \\ $$$$\left.\:\:\:+\mathrm{2}\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}\right)\left(\mathrm{16}\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\mathrm{8}\boldsymbol{\mathrm{ax}}^{\mathrm{2}} \right) \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{256}+\boldsymbol{\mathrm{a}}\right)\boldsymbol{\mathrm{x}}^{\mathrm{6}} −\mathrm{16}\boldsymbol{\mathrm{x}}^{\mathrm{5}} +\left(\mathrm{64}\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\mathrm{32}\boldsymbol{\mathrm{b}}+\mathrm{8}\boldsymbol{\mathrm{a}}\right)\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\mathrm{8}\left(\boldsymbol{\mathrm{a}}+\mathrm{1}\right)\boldsymbol{\mathrm{x}}^{\mathrm{3}} \\ $$$$\left(\mathrm{32}\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\mathrm{16}\boldsymbol{\mathrm{ab}}−\mathrm{8}\right)\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{a}}+\mathrm{1}\right)\boldsymbol{\mathrm{x}}+\mathrm{16}\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\mathrm{a}^{\mathrm{4}} −\boldsymbol{\mathrm{a}}\left(\mathrm{1}−\mathrm{2}\boldsymbol{\mathrm{b}}\right)=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{256}\left(\mathrm{1}+\boldsymbol{\mathrm{a}}\right)\boldsymbol{\mathrm{x}}^{\mathrm{6}} −\mathrm{16}\boldsymbol{\mathrm{x}}^{\mathrm{5}} +\mathrm{8}\left(\mathrm{8a}^{\mathrm{2}} +\mathrm{4b}+\mathrm{a}\right)\boldsymbol{\mathrm{x}}^{\mathrm{4}} \\ $$$$+\mathrm{8}\left(\boldsymbol{\mathrm{a}}+\mathrm{1}\right)\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{8}\left(\mathrm{4}\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{ab}}−\mathrm{1}\right)\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\left(\boldsymbol{\mathrm{a}}+\mathrm{1}\right)^{\mathrm{2}} \boldsymbol{\mathrm{x}} \\ $$$$+\boldsymbol{\mathrm{a}}^{\mathrm{4}} +\mathrm{16}\boldsymbol{\mathrm{a}}^{\mathrm{3}} −\boldsymbol{\mathrm{a}}+\mathrm{2}\boldsymbol{\mathrm{ab}}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$\left(\mathrm{256}\boldsymbol{\mathrm{x}}^{\mathrm{6}} −\mathrm{16}\boldsymbol{\mathrm{x}}^{\mathrm{5}} +\mathrm{8}\boldsymbol{\mathrm{x}}^{\mathrm{3}} −\mathrm{8}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}\right)+ \\ $$$$\mathrm{256}\boldsymbol{\mathrm{ax}}^{\mathrm{6}} +\mathrm{8}\left(\mathrm{8}\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{a}}\right)\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\mathrm{8}\boldsymbol{\mathrm{ax}}^{\mathrm{3}} \\ $$$$+\mathrm{16}\left(\mathrm{2}\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\boldsymbol{\mathrm{ab}}\right)\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{a}}\right)\boldsymbol{\mathrm{x}}+ \\ $$$$\boldsymbol{\mathrm{a}}^{\mathrm{4}} +\mathrm{16}\boldsymbol{\mathrm{a}}^{\mathrm{3}} −\boldsymbol{\mathrm{a}}+\mathrm{2}\boldsymbol{\mathrm{ab}}=\mathrm{0} \\ $$$$ \\ $$$$\begin{cases}{\mathrm{256}\boldsymbol{\mathrm{x}}^{\mathrm{5}} −\mathrm{16}\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\mathrm{8}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{\mathrm{x}}−\mathrm{1}=\mathrm{0}}\\{\mathrm{8}\left[\mathrm{32}\boldsymbol{\mathrm{ax}}^{\mathrm{6}} +\left(\mathrm{8}\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{a}}\right)\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\boldsymbol{\mathrm{ax}}^{\mathrm{3}} +\right.}\end{cases} \\ $$$$\left.\boldsymbol{\mathrm{a}}\left(\mathrm{4}\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\right)\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\left(\boldsymbol{\mathrm{a}}+\mathrm{2}\right)\boldsymbol{\mathrm{x}}\right]+ \\ $$$$\boldsymbol{\mathrm{a}}\left(\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\mathrm{26}\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\mathrm{1}+\mathrm{2}\boldsymbol{\mathrm{b}}\right)=\mathrm{0} \\ $$$$ \\ $$$$\:\:\begin{cases}{\boldsymbol{\mathrm{x}}=\mathrm{0},\mathrm{112166986}…\:\:\mathrm{and}\:\mathrm{imsg}\left(\mathrm{x}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right)}\\{\mathrm{8}\boldsymbol{\mathrm{ax}}\left(\mathrm{32}\boldsymbol{\mathrm{x}}^{\mathrm{5}} +\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\left(\mathrm{4}\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\right)\boldsymbol{\mathrm{x}}−\left(\boldsymbol{\mathrm{a}}+\mathrm{2}\right)=\mathrm{0}\left(\mathrm{2}\right)\right.}\end{cases} \\ $$$$\bullet\mathrm{3}\:\:\:\mathrm{a}^{\mathrm{3}} +\mathrm{16a}^{\mathrm{2}} +\mathrm{2b}−\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{1}\bullet\:\:\:\:\mathrm{x}= \\ $$$$\mathrm{2}\boldsymbol{\mathrm{et}}\:\mathrm{3}\bullet\:\:\:\:\begin{cases}{\mathrm{32}\boldsymbol{\mathrm{x}}^{\mathrm{5}} +\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\left(\mathrm{4}\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\right)\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{a}}+\mathrm{2}}\\{\:\:\:\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\mathrm{16}\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{b}}−\mathrm{1}\:\:\:\:\:=\mathrm{0}}\end{cases} \\ $$$$ \\ $$$$\mathrm{a}^{\mathrm{2}} \left(\mathrm{a}+\mathrm{16}\right)=\mathrm{1}−\mathrm{2b} \\ $$$$\:\:\:\boldsymbol{\mathrm{b}}=\frac{\mathrm{1}−\boldsymbol{\mathrm{a}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{a}}+\mathrm{16}\right)}{\mathrm{2}} \\ $$$$\mathrm{4}\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}=\frac{\boldsymbol{\mathrm{a}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{a}}+\mathrm{16}\right)+\mathrm{8}\boldsymbol{\mathrm{a}}−\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\Rightarrow\:\:\mathrm{32x}^{\mathrm{5}} +\mathrm{x}^{\mathrm{2}} +\frac{\boldsymbol{\mathrm{a}}^{\mathrm{2}} \left(\mathrm{a}+\mathrm{16}\right)+\mathrm{8}\boldsymbol{\mathrm{a}}−\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{a}}+\mathrm{2} \\ $$$$\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{change}}\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{by}}\:\:\boldsymbol{\mathrm{x}}_{\mathrm{0}} =\mathrm{0},\mathrm{1121167}. \\ $$$$\Rightarrow\mathrm{we}\:\mathrm{bave}\:\:\mathrm{equation}\:\:\:\mathrm{with}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{as}}\:\boldsymbol{\mathrm{unconnu}} \\ $$$$\boldsymbol{\mathrm{then}}\:\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{found}}\:\:\:\boldsymbol{\mathrm{a}}\:\:\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{b}}\: \\ $$$$\left(\mathrm{exept}\:\mathrm{Ereur}\:\mathrm{in}\:\mathrm{the}\:\mathrm{Calcul}\right) \\ $$
Commented by Ghisom last updated on 03/Sep/24
$$\mathrm{some}\:\mathrm{steps}\:\mathrm{are}\:\mathrm{not}\:\mathrm{clear}\:\mathrm{and}\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think} \\ $$$$\mathrm{your}\:\mathrm{result}\:\mathrm{holds}. \\ $$