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Question Number 211255 by ajfour last updated on 02/Sep/24
(√(a+(√(b−x))+(√(b−(√(a+x))))))=2x solve for x.
a+b−x+b−a+x=2xsolveforx.
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]
a+b−x+b−a+x=2xx=(b−(4x2−b−x−a)2)2−ait′salreadyof8thdegreeandwestillhaveb−x⇒it′sof16thdegree.wecan′tsolveforxnorfora.butwecanforb:b=256x8−256ax6+32x5+96a2x4−16ax3−(16a3−1)x2+(2a2+1)x+a(a3+1)4(4x2−a)2+(16x4−8ax2−x+a2)a+x2(4x2−a)2[mightnotalwaysbetruebecauseof3timessquaring]
Commented by ajfour last updated on 06/Sep/24
(√(a+(√(b−x))))+(√(b−(√(a+x))))=2x i think i was interested in this! rather
a+b−x+b−a+x=2xithinkiwasinterestedinthis!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 .....
leta+x=p2,b−x=q2p2+q2=a+b2x=(p2−q2)−(a−b)a+q+b−p=p2−q2−(a−b)squaringa+q={2(p2−a)−b−p}2a+b−p2=[{2(p2−a)−b−p}2−a]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)
4x2−a=b−x+b−a+x(4x2−a)−b−x=b−a+x(4x2−a)2−x−2(4x2−a)b−x=−a+x((4x2−a)2−x=2(4x2−a)b−x−a+x4x2−a)2+a+x=2(4x2−a)b−x4x2−a2+a+x2(4x2−a)=b−x(4x2−a)24+a+x4(4x2−a)2+(4x2−a)a+x2=b−xa+x=2(4x2−a)[b−x−a+x4(4x2−a)2−(4x2−a)24]a+x=24x2−a[b−x−([(a+x)+(4x2−a)44(4x2−a)2)=[2(b−x)(4x2−a)−(a+x)−(4x2−a)4(4x2−a)2]=[(4x2−a)[(b−x)−(4x2−a)3](4x2−a)2a+x=(4x2−a)2(b−x−(4x2−a)3)2(4x2−a)4(4x2−a)4(a+x)=(4x2−a)2[(b−x)2+(4x2−a)6−2(b−x)(4x2−a)3](4x2−a)2(a+x)=(b−x)2+(4x2−a)6−2(b−x)(4x2−a)3](4x2−a)2(a+x)−(4x2−a)6+2(b−x)(4x2−a)3=(b−x)2(4x2−a)2[a+x−(4x2−a)4+2(b−x)(4x2−a)]=(b−x)2[a+x+8bx2−2ab−8x3+2ax−(4x2−a)4=(b−x)2(b−x)2+(4x2−a)4=[(b−x+(4x2−a)2]2−2(b−x)(4x2−a)2soit−8x3+8bx2+(2a+1)x+a(1−2b)=b2+x2−2bx+(16x4+a2−8ax2)2+2(b−x)(4x2−a)2=x2−2bx+b2+[8x2(a+2x2)+a2]+64x4(a+2x2)2+a4+16a2(a+2x2)+2b−x)(16x4+a2−8ax2)⇒(256+a)x6−16x5+(64a2+32b+8a)x4+8(a+1)x3(32a2−16ab−8)x2−(a2+2a+1)x+16a3+a4−a(1−2b)=0256(1+a)x6−16x5+8(8a2+4b+a)x4+8(a+1)x3+8(4a2−2ab−1)x2−(a+1)2x+a4+16a3−a+2ab=0(256x6−16x5+8x3−8x2−x)+256ax6+8(8a2+4b+a)x4+8ax3+16(2a2−ab)x2−(a2+2a)x+a4+16a3−a+2ab=0{256x5−16x4+8x2+8x−1=08[32ax6+(8a2+4b+a)x4+ax3+a(4a−b)x2−a(a+2)x]+a(a3+26a2−1+2b)=0{x=0,112166986…andimsg(x)(1)8ax(32x5+x2+(4a−b)x−(a+2)=0(2)∙3a3+16a2+2b−1=01∙x=2et3∙{32x5+x2+(4a−b)x=a+2a3+16a2+2b−1=0a2(a+16)=1−2bb=1−a2(a+16)24a−b=a2(a+16)+8a−12(2)⇒32x5+x2+a2(a+16)+8a−12x=a+2wechangexbyx0=0,1121167.⇒webaveequationwithaasunconnuthenwefoundaandb(exeptEreurintheCalcul)
Commented by Ghisom last updated on 03/Sep/24
some steps are not clear and I don′t think your result holds.
somestepsarenotclearandIdon′tthinkyourresultholds.

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