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Question-209385

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Question Number 209385 by efronzo1 last updated on 08/Jul/24
Answered by Rasheed.Sindhi last updated on 08/Jul/24
p(x)=ax^2 +bx+1⇒p(1)=a+b+1 q(x)=bx^2 +ax+1⇒q(1)=b+a+1 p( q(1) )=q( p(1) ) p(a+b+1)=q(a+b+1) a(a+b+1)^2 +b(a+b+1)+1 =b(a+b+1)^2 +a(a+b+1)+1 a(a^2 +b^2 +1+2ab+2b+2a)+ab+b^2 +b =b(a^2 +b^2 +1+2ab+2b+2a)+a^2 +ab+a a^3 +ab^2 +a+2a^2 b+2ab+2a^2 +b^2 +b =a^2 b+b^3 +b+2ab^2 +2b^2 +2ab+a^2 +a a^3 −b^3 −ab^2 +a^2 b+a−b+a^2 −b^2 +a−b=0 a^3 −b^3 −ab^2 +a^2 b+a^2 −b^2 =0 (a−b)(a^2 +ab+b^2 )+ab(a−b)+(a−b)(a+b)=0 (a−b)(a^2 +ab+b^2 +ab+a+b)=0 a−b≠0⇒(a+b)^2 +(a+b)=0 (a+b)(a+b+1)=0 a+b≠0⇒a+b=−1 p(1)=0 , q(1)=0 p(0)=1, q(0)=1 ▶q(q(q(0)))+p(p(p(1))) =q(q(1))+p(p(0)) =q(0)+p(1) =1+0=1
p(x)=ax2+bx+1⇒p(1)=a+b+1q(x)=bx2+ax+1⇒q(1)=b+a+1p(q(1))=q(p(1))p(a+b+1)=q(a+b+1)a(a+b+1)2+b(a+b+1)+1=b(a+b+1)2+a(a+b+1)+1a(a2+b2+1+2ab+2b+2a)+ab+b2+b=b(a2+b2+1+2ab+2b+2a)+a2+ab+aa3+ab2+a+2a2b+2ab+2a2+b2+b=a2b+b3+b+2ab2+2b2+2ab+a2+aa3−b3−ab2+a2b+a−b+a2−b2+a−b=0a3−b3−ab2+a2b+a2−b2=0(a−b)(a2+ab+b2)+ab(a−b)+(a−b)(a+b)=0(a−b)(a2+ab+b2+ab+a+b)=0a−b≠0⇒(a+b)2+(a+b)=0(a+b)(a+b+1)=0a+b≠0⇒a+b=−1p(1)=0,q(1)=0p(0)=1,q(0)=1▸q(q(q(0)))+p(p(p(1)))=q(q(1))+p(p(0))=q(0)+p(1)=1+0=1
Commented by Frix last updated on 08/Jul/24
Yes but take a closer look: a(a+b+1)^2 +b(a+b+1)+1 =b(a+b+1)^2 +a(a+b+1)+1 it directly follows: a(a+b+1)+b=b(a+b+1)+a (a−b)(a+b+1)=a−b a+b+1=0
Yesbuttakeacloserlook:a(a+b+1)2+b(a+b+1)+1=b(a+b+1)2+a(a+b+1)+1itdirectlyfollows:a(a+b+1)+b=b(a+b+1)+a(a−b)(a+b+1)=a−ba+b+1=0
Commented by Rasheed.Sindhi last updated on 09/Jul/24
Yes sir, thanks!
Yessir,thanks!

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