Question Number 100193 by bobhans last updated on 25/Jun/20

(√(7+2(√(7−2(√(7+2(√(7−2(√(7+...)))))))))) ?

Commented bybobhans last updated on 25/Jun/20

x =(√(7+2y)) ; y=(√(7−2x)) ⇔x^2 = 7+2y _(1) ⇔y^2 = 7−2x _(2) ⇒ (((x^2 −7)/2))^2 = 7−2x x^4 −14x^2 +49 = 28−8x x^4 −14x^2 +8x+21 = 0 ⇒ x = 3 ∧ y = 1 ; then x=(√(7+2(√(7−2(√(7+2(√(7−...)))))))) x =(√(7+2.1)) = 3 ■

Commented byDwaipayan Shikari last updated on 25/Jun/20

Yes it is 3

Commented byDwaipayan Shikari last updated on 25/Jun/20

I also did this

Answered by Dwaipayan Shikari last updated on 25/Jun/20

Suppose it is p (√(7+2(√(7−2(√(7+2(√(7−2..))))))))∞=p 7+2(√(7−2(√(7+2(√(7−2(√(7+2..))))))))=p^2 7−2(√(7+2(√(7−2(√(7+2(√(7−2...))))))))=(((p^2 −7)/2))^2 (√(7+2(√(7−2(√(7+2(√(7−2...))))))))=((28−(p^2 −7)^2 )/8)=p p=((28−(p^2 −7)^2 )/8)=((28−p^4 +14p^2 −49)/8) so,p^4 −14p^2 +8p+21=0 p=3■