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Question Number 50511 by behi83417@gmail.com last updated on 17/Dec/18

refer to question #50278  (√(a+(√(a−x))))+(√(a−(√(a+x))))=2x

You can't use 'macro parameter character #' in math modea+ax+aa+x=2x

Answered by behi83417@gmail.com last updated on 17/Dec/18

Commented by behi83417@gmail.com last updated on 17/Dec/18

if a=0,then: x=0 can accept as answer.

ifa=0,then:x=0canacceptasanswer.

Commented by behi83417@gmail.com last updated on 17/Dec/18

Commented by behi83417@gmail.com last updated on 17/Dec/18

Commented by behi83417@gmail.com last updated on 17/Dec/18

Commented by mr W last updated on 17/Dec/18

can you give the solution for a  concrete value of a, e.g. a=20?

canyougivethesolutionforaconcretevalueofa,e.g.a=20?

Commented by behi83417@gmail.com last updated on 17/Dec/18

excuse me sir.i cant undrestand this  commenet.   [concrete=?]

excusemesir.icantundrestandthiscommenet.[concrete=?]

Commented by mr W last updated on 17/Dec/18

I mean you take a special value for the  parameter a, say a=20, and show  your solution for this special case.  i.e.  (√(20+(√(20−x))))+(√(20−(√(20+x))))=2x

Imeanyoutakeaspecialvaluefortheparametera,saya=20,andshowyoursolutionforthisspecialcase.i.e.20+20x+2020+x=2x

Commented by behi83417@gmail.com last updated on 18/Dec/18

a=20,is too large and big numbers in  solution.solve for:a=1  (√(1+(√(1−x))))+(√(1−(√(1+x))))=2x  1+x=t^2 ,1−x=s^2 ⇒x=((t^2 −s^2 )/2),t^2 +s^2 =2  (√(1+s))+(√(1−t))=t^2 −s^2   1+s+1−t+2(√((1+s)(1−t)))=(t^2 −s^2 )^2   4(1+s)(1−t)=[(t^2 −s^2 )^2 +t−s−2]^2   4−4t+4s−4ts=[t^4 +s^4 −2t^2 s^2 +t−s−2]^2 =  =t^8 +s^8 +4t^4 s^4 +t^2 +s^2 +4+2t^4 s^4 −4t^6 s^2   +2t^5 −2t^4 s−4t^4 −4t^2 s^6 +2ts^4 −2s^5 −4s^4   −4t^3 s^2 +4t^2 s^3 +4t^2 s^2 −4t+4s−2ts  (t^8 +s^8 )+2(t^5 −s^5 )+6t^4 s^4 +2ts+4t^2 s^2   −4t^2 s^2 (t^4 +s^4 )−2ts(t^3 −s^3 )+4(t^4 +s^4 )  −4t^2 s^2 (t−s)=0  (t^8 +s^8 )+2(t^5 −s^5 )−4(1+t^2 s^2 )(t^4 +s^4 )  −2ts(t^3 −s^3 )−4t^2 s^2 (t−s)+6t^4 s^4 +  +8t^2 s^2 +2ts+(t^2 +s^2 )=0  let:      t−s=p   ,   ts=q  t^8 +s^8 =p^8 +8p^6 q+20p^4 q^2 +16p^2 q^3 +2q^4   2(t^5 −s^5 )=2p^5 +10p^2 q+10pq^2   4(1−t^2 s^2 )(t^4 +s^4 )=4(p^4 +4p^2 q+2q^2 )(1−q^2 )=  =4p^4 +16p^2 q+8q^2 −4p^4 q^2 −16p^2 q^3 −8q^4   −2ts(t^3 −s^3 )=−2q(p^3 +3pq)=  =−2p^3 q−6pq^2   −4t^2 s^2 (t−s)=−4q^2 p  −2p^3 q−10pq^2 +2q+4p^4 =0  16p^4 q^2 +2(4p^6 +4p^3 −8p^2 +2)q  +(p^8 +2p^5 −4p^4 +p^2 )=0  △′=(4p^6 +4p^3 −8p^2 +2)^2 −16p^4 (p^8 +2p^5 −4p^4 +p^2 )=  =4(2p^3 −4p^2 +1)^2   q=((−(4p^6 +4p^3 −8p^2 +2)±2(2p^3 −4p^2 +1))/(16p^4 ))=  q_1 =((−4p^6 )/(16p^4 ))=−(p^2 /4)  q_2 =−((p^6 +2p^3 −4p^2 +1)/(4p^4 ))  for:q=−(p^2 /4)⇒p^2 +4q=0⇒(t−s)^2 +4ts=0  ⇒(t+s)^2 =0⇒t=−s⇒x=0 (not ok!)  q(p)=−((p^6 +2p^3 −4p^2 +1)/(4p^4 )),have extrimums  at points: E(−.64,1.63),F(.75,.18)  p=−.64⇒t−s=−0.64⇒t=s−0.64  q=1.63⇒ts=1.63⇒s(s−.64)=1.63  ⇒s^2 −0.64s−1.63=0⇒s=((.64±2.63)/2)  ⇒s=1.64,−1⇒t=1,−1.64  ⇒x=((t^2 −s^2 )/2)=((1.64^2 −1)/2)=±0.84  (√(1+(√(1−0.84))))+(√(1−(√(1+0.84))))=  =1.18+0.06i≠2×0.84  (√(1+(√(1+.84))))+(√(1−(√(1−.84))))≠2(−.84)  ⇒x=±0.84 not answer.  p=0.75⇒t−s=0.75⇒t=s+0.75  ts=.18⇒s(s+.75)=.18  ⇒s^2 +.75s−.18=0⇒s=((−.75±1.13)/2)  ⇒s=−.94,0.19  ,  t=−.19,0.94  x=(((.94)^2 −(.19)^2 )/2)=±0.42  (√(1+(√(1−.42))))+(√(1−(√(1+.42))))≠2×(.42)  (√(1+(√(1+.42))))+(√(1−(√(1−.42))))≠2(−.42)  ⇒x=±.42   not  answer.

a=20,istoolargeandbignumbersinsolution.solvefor:a=11+1x+11+x=2x1+x=t2,1x=s2x=t2s22,t2+s2=21+s+1t=t2s21+s+1t+2(1+s)(1t)=(t2s2)24(1+s)(1t)=[(t2s2)2+ts2]244t+4s4ts=[t4+s42t2s2+ts2]2==t8+s8+4t4s4+t2+s2+4+2t4s44t6s2+2t52t4s4t44t2s6+2ts42s54s44t3s2+4t2s3+4t2s24t+4s2ts(t8+s8)+2(t5s5)+6t4s4+2ts+4t2s24t2s2(t4+s4)2ts(t3s3)+4(t4+s4)4t2s2(ts)=0(t8+s8)+2(t5s5)4(1+t2s2)(t4+s4)2ts(t3s3)4t2s2(ts)+6t4s4++8t2s2+2ts+(t2+s2)=0let:ts=p,ts=qt8+s8=p8+8p6q+20p4q2+16p2q3+2q42(t5s5)=2p5+10p2q+10pq24(1t2s2)(t4+s4)=4(p4+4p2q+2q2)(1q2)==4p4+16p2q+8q24p4q216p2q38q42ts(t3s3)=2q(p3+3pq)==2p3q6pq24t2s2(ts)=4q2p2p3q10pq2+2q+4p4=016p4q2+2(4p6+4p38p2+2)q+(p8+2p54p4+p2)=0=(4p6+4p38p2+2)216p4(p8+2p54p4+p2)==4(2p34p2+1)2q=(4p6+4p38p2+2)±2(2p34p2+1)16p4=q1=4p616p4=p24q2=p6+2p34p2+14p4for:q=p24p2+4q=0(ts)2+4ts=0(t+s)2=0t=sx=0(notok!)q(p)=p6+2p34p2+14p4,haveextrimumsatpoints:E(.64,1.63),F(.75,.18)p=.64ts=0.64t=s0.64q=1.63ts=1.63s(s.64)=1.63s20.64s1.63=0s=.64±2.632s=1.64,1t=1,1.64x=t2s22=1.64212=±0.841+10.84+11+0.84==1.18+0.06i2×0.841+1+.84+11.842(.84)x=±0.84notanswer.p=0.75ts=0.75t=s+0.75ts=.18s(s+.75)=.18s2+.75s.18=0s=.75±1.132s=.94,0.19,t=.19,0.94x=(.94)2(.19)22=±0.421+1.42+11+.422×(.42)1+1+.42+11.422(.42)x=±.42notanswer.

Commented by mr W last updated on 18/Dec/18

what is the solution? i think for a=1  there is no solution.

whatisthesolution?ithinkfora=1thereisnosolution.

Commented by behi83417@gmail.com last updated on 18/Dec/18

for:a=1, no answer excist.

for:a=1,noanswerexcist.

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