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Question Number 50278 by LYCON TRIX last updated on 15/Dec/18
(√(a+(√(a−x)))) + (√(a−(√(a+x)))) = 2x please i beg u guys please solve this question
a+a−x+a−a+x=2xpleaseibeguguyspleasesolvethisquestion
Commented by ajfour last updated on 15/Dec/18
a and x ∈ R or C ?
aandx∈RorC?
Commented by MJS last updated on 15/Dec/18
would you be so kind to tell us where this equation comes from?
wouldyoubesokindtotelluswherethisequationcomesfrom?
Commented by maxmathsup by imad last updated on 15/Dec/18
its only a trying let use the changement x =acos(2t) (e) ⇔ (√(a+(√(a−acos(2t)))))+(√(a−(√a)))=2a cos(2t) ⇒ (√(a+(√a)(√(1−cos(2t)))))+(√(a−(√a)(√(1+cos(2t)))))=2a cos(2t) ⇒ (√(a+(√(2a))sint))+(√(a−(√(2a))cost))=2acos(2t) ⇒ a+(√(2a))sint +2(√(a+(√(2a))sint))(√(a−(√(2a))cost)) +a−(√(2a))cost =4a^2 cos^2 (2t) ⇒ 2a+(√(2a))(sint −cost) +2(√((a+(√(2a))sint)(a−(√(2a))cost))) =4a^2 cos^2 (2t) ⇒ (4a^2 cos^2 (2t)−2a−(√(2a))(sint −cost))^2 =4(a^2 −a(√(2a))cost +a(√(2a))sint −2asint cost) ....be cntinued i think that way can give something if we resist to storm of calculus...
itsonlyatryingletusethechangementx=acos(2t)(e)⇔a+a−acos(2t)+a−a=2acos(2t)⇒a+a1−cos(2t)+a−a1+cos(2t)=2acos(2t)⇒a+2asint+a−2acost=2acos(2t)⇒a+2asint+2a+2asinta−2acost+a−2acost=4a2cos2(2t)⇒2a+2a(sint−cost)+2(a+2asint)(a−2acost)=4a2cos2(2t)⇒(4a2cos2(2t)−2a−2a(sint−cost))2=4(a2−a2acost+a2asint−2asintcost)….becntinuedithinkthatwaycangivesomethingifweresisttostormofcalculus…
Commented by ajfour last updated on 23/Aug/23
A past prize question, i wonder who got the prize. just name the guy, wont u?
Apastprizequestion,iwonderwhogottheprize.justnametheguy,wontu?
Commented by ajfour last updated on 15/Mar/25
okay, i now know how much i need. ..say 50 crore rupees ( Indian currency) would amply suffice.
okay,inowknowhowmuchineed...say50crorerupees(Indiancurrency)wouldamplysuffice.
Commented by ajfour last updated on 15/Mar/25
Answered by behi83417@gmail.com last updated on 17/Dec/18
a+x=t^2 ,a−x=s^2 ⇒x=((t^2 −s^2 )/2),a=((t^2 +s^2 )/2) (√(a+s))+(√(a−t))=t^2 −s^2 a+s+a−t+2(√((a+s)(a−t)))=(t^2 −s^2 )^2 2(√((a+s)(a−t)))=(t^2 −s^2 )^2 +t−s−2a= =t^4 +s^4 −2t^2 s^2 −t^2 −s^2 +t−s ⇒4(a^2 −at+as−st)= =t^8 +s^8 +4t^4 s^4 +t^4 +s^4 +t^2 +s^2 + +2t^4 s^4 −4t^6 s^2 −2t^6 −2t^4 s^2 +2t^5 −2t^4 s −4t^2 s^6 −2s^4 t^2 −2s^6 +2ts^4 −2s^5 +4t^4 s^2 +4t^2 s^4 −4t^3 s^2 +4t^2 s^3 +2t^2 s^2 −2t^3 +2t^2 s −2s^2 t+2s^3 −2ts LHS=(t^2 +s^2 )^2 −2(t^2 +s^2 )(t−s)−4st= =t^4 +s^4 +2t^2 s^2 −2t^3 +2t^2 s−2s^2 t+2s^3 −4st RHS=t^8 +s^8 +6t^4 s^4 +t^4 +s^4 +t^2 +s^2 −4t^6 s^2 −2t^6 −2s^6 +2t^4 s^2 +2t^5 −2t^4 s −4t^2 s^6 −2s^4 t^2 +2ts^4 −2s^5 +4t^2 s^4 −4t^3 s^2 +4t^2 s^3 +2t^2 s^2 −2t^3 +2t^2 s−2s^2 t+2s^3 −2ts ⇒t^8 +s^8 −2t^6 −2s^6 +2t^5 +2s^5 +t^2 +s^2 +2ts−4t^6 s^2 +2t^4 s^2 −2t^4 s−4t^2 s^6 +2s^4 t^2 +2ts^4 −4t^3 s^2 +4t^2 s^3 +2t^2 s −2s^2 t=0 (t^8 +s^8 )−2(t^6 +s^6 )+2(t^5 −s^5 )− −4t^2 s^2 (t^4 +s^4 )−2ts(t^3 −s^3 )+ +(2t^2 s^2 +1)(t^2 +s^2 )−4t^2 s^2 (t−s)+ +6t^4 s^4 +2ts=0 t−s=p,ts=q t^8 +s^8 =(t^4 −s^4 )^2 +2t^4 s^4 = =(t^2 −s^2 )^2 (t^2 +s^2 )^2 +2t^4 s^4 = =(t−s)^2 (t+s)^2 (t^2 +s^2 )^2 +2t^4 s^4 = =(t−s)^2 ((t−s)^2 +4ts)((t−s)^2 +2ts)^2 +2t^4 s^4 = =(p)^2 (p^2 +4q)(p^4 +4p^2 q+4q^2 )+2q^4 = =(p^4 +4p^2 q)(p^4 +4p^2 q+4q^2 )+2q^4 = =p^8 +8p^6 q+20p^4 q^2 +16p^2 q^3 +2q^4 t^6 +s^6 =(t^3 −s^3 )^2 +2t^3 s^3 =(t−s)^2 (t^2 +ts+s^2 )^2 +2t^3 s^3 = =(p)^2 ((t−s)^2 +3ts)^2 +2t^3 s^3 = =p^2 (p^2 +3q)^2 +2q^3 = =p^6 +6p^4 q+9p^2 q^2 +2q^3 t^5 −s^5 =(t−s)(t^4 +t^3 s+t^2 s^2 +ts^3 +s^4 )= =(t−s)(t^4 +s^4 +ts(t^2 +ts+s^2 ))= =(t−s)((t^2 +s^2 )^2 −2t^2 s^2 +ts((t−s)^2 +3ts)= =(t−s)[((t−s)^2 +2ts)^2 −2t^2 s^2 +ts((t−s)^2 +3ts))= =p[((p^2 +2q)^2 −2q^2 +q(p^2 +3q))= =p[(p^4 +4p^2 q+4q^2 )−2q^2 +qp^2 +3pq]= =p^5 +5p^2 q+5pq^2 t^4 +s^4 =(t^2 +s^2 )^2 −2t^2 s^2 =[(t−s)^2 +2ts]^2 −2t^2 s^2 = =[p^2 +2q]^2 −2q^2 =p^4 +4p^2 q+2q^2 t^3 −s^3 =(t−s)(t^2 +s^2 +ts)=p^3 +3pq t^2 +s^2 =(t−s)^2 +2ts=p^2 +2q after replacing and simplifing: (p^8 −2p^6 +2p^5 +p^2 )+16p^2 (p^2 −1)q^2 +(8p^6 −12p^4 +8p^3 +4)q=0
a+x=t2,a−x=s2⇒x=t2−s22,a=t2+s22a+s+a−t=t2−s2a+s+a−t+2(a+s)(a−t)=(t2−s2)22(a+s)(a−t)=(t2−s2)2+t−s−2a==t4+s4−2t2s2−t2−s2+t−s⇒4(a2−at+as−st)==t8+s8+4t4s4+t4+s4+t2+s2++2t4s4−4t6s2−2t6−2t4s2+2t5−2t4s−4t2s6−2s4t2−2s6+2ts4−2s5+4t4s2+4t2s4−4t3s2+4t2s3+2t2s2−2t3+2t2s−2s2t+2s3−2tsLHS=(t2+s2)2−2(t2+s2)(t−s)−4st==t4+s4+2t2s2−2t3+2t2s−2s2t+2s3−4stRHS=t8+s8+6t4s4+t4+s4+t2+s2−4t6s2−2t6−2s6+2t4s2+2t5−2t4s−4t2s6−2s4t2+2ts4−2s5+4t2s4−4t3s2+4t2s3+2t2s2−2t3+2t2s−2s2t+2s3−2ts⇒t8+s8−2t6−2s6+2t5+2s5+t2+s2+2ts−4t6s2+2t4s2−2t4s−4t2s6+2s4t2+2ts4−4t3s2+4t2s3+2t2s−2s2t=0(t8+s8)−2(t6+s6)+2(t5−s5)−−4t2s2(t4+s4)−2ts(t3−s3)++(2t2s2+1)(t2+s2)−4t2s2(t−s)++6t4s4+2ts=0t−s=p,ts=qt8+s8=(t4−s4)2+2t4s4==(t2−s2)2(t2+s2)2+2t4s4==(t−s)2(t+s)2(t2+s2)2+2t4s4==(t−s)2((t−s)2+4ts)((t−s)2+2ts)2+2t4s4==(p)2(p2+4q)(p4+4p2q+4q2)+2q4==(p4+4p2q)(p4+4p2q+4q2)+2q4==p8+8p6q+20p4q2+16p2q3+2q4t6+s6=(t3−s3)2+2t3s3=(t−s)2(t2+ts+s2)2+2t3s3==(p)2((t−s)2+3ts)2+2t3s3==p2(p2+3q)2+2q3==p6+6p4q+9p2q2+2q3t5−s5=(t−s)(t4+t3s+t2s2+ts3+s4)==(t−s)(t4+s4+ts(t2+ts+s2))==(t−s)((t2+s2)2−2t2s2+ts((t−s)2+3ts)==(t−s)[((t−s)2+2ts)2−2t2s2+ts((t−s)2+3ts))==p[((p2+2q)2−2q2+q(p2+3q))==p[(p4+4p2q+4q2)−2q2+qp2+3pq]==p5+5p2q+5pq2t4+s4=(t2+s2)2−2t2s2=[(t−s)2+2ts]2−2t2s2==[p2+2q]2−2q2=p4+4p2q+2q2t3−s3=(t−s)(t2+s2+ts)=p3+3pqt2+s2=(t−s)2+2ts=p2+2qafterreplacingandsimplifing:(p8−2p6+2p5+p2)+16p2(p2−1)q2+(8p6−12p4+8p3+4)q=0
Commented by behi83417@gmail.com last updated on 17/Dec/18
⇒Q#50511
You can't use 'macro parameter character #' in math mode

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