Question Number 103321 by bemath last updated on 14/Jul/20

5+(√(5−(√(5+(√(5−(√(5+(√(5−(√(5+...))))))))))))

Answered by Dwaipayan Shikari last updated on 14/Jul/20

(√(5−(√(5+(√(5...))))))=p  5−(√(5+(√(5−(√5)))))...=p^2   5+p=(5−p^2 )^2   p^4 −10p^2 +20−p=0  (p^2 +p−5)(p^2 −p−4)=0  p=((−1±(√(21)))/2)  or p=((1±(√(17)))/2)  5−((1±(√(21)))/2)   or  5+((1±(√(17)))/2)  More precisely  p=((10−1+(√(17)))/2)=((9+(√(17)))/2)(A possible one)

Answered by OlafThorendsen last updated on 14/Jul/20

x = 5+(√(5−(√(5+(√(5−(√(5+(√(5−(√(5+...))))))))))))  x = 5+(√(5−(√x)))  ⇒ (x−5)^2  = 5−(√x)  x^2 −10x+(√x)+20 = 0  x = X^2   X^4 −10X^2 +X+20 = 0   { ((X = (1/2)−((√(17))/2) (1))),((X = (1/2)+((√(17))/2) (2))),((X = −(1/2)−((√(21))/2) (3))),((X = −(1/2)+((√(21))/2) (4))) :}  (1) and (3) impossible (X>0)  x>5 ⇒ X>(√5) ≈ 2,23  ⇒ (4) impossible  Then X = (1/2)+((√(17))/2)  And x = ((1+2(√(17))+17)/4) = ((9+(√(17)))/2)