Question Number 103914 by bobhans last updated on 18/Jul/20

(2+(√3))^x^2   + (2−(√3))^x^2   = 4

Commented bysom(math1967) last updated on 18/Jul/20

let (2+(√3))^x^2  =a  ∴(2−(√3))^x^2  =(1/a)  ⇒a^2 −4a+1=0  ⇒a=(2±(√3))  (2+(√3))^x^2  =(2+(√3))  ⇒x=±1  (2+(√3))^x^2  =(2+(√3))^(−1)   x=i

Commented bybemath last updated on 18/Jul/20

thank you both

Answered by OlafThorendsen last updated on 18/Jul/20

X = (2+(√3))^x^2    (2−(√3))^x^2   = ((1/(2+(√3))))^x^2  = (1/((2+(√3))^x^2  )) = (1/X)  then X+(1/X) = 4  X^2 −4X+1 = 0  Δ = (−4)^2 −4(1)(1) = 12  X = ((−(−4)±(√(12)))/(2(1))) = 2±(√3)  1) X = 2+(√3)  x^2  = 1, x=±1  2) X = 2−(√3)  2−(√3) = (2+(√3))^x^2    (1/(2+(√3))) = (2+(√3))^x^2    (2+(√3))^(x^2 +1)  = 1  x^2 +1 = 0  x = ±i  S = {±1;±i}

Answered by Rauny last updated on 18/Jul/20

(2+(√3))^x^2   + (2−(√3))^x^2   = 4   t:=2+(√3)  (1/(2+(√3)))=((2−(√3))/((2+(√3))(2−(√3))))=2−(√3)  ∴2−(√3)=(1/t)  t^x^2  +t^(−x^2 ) −4=0  u:=t^x^2   (∀u>0)  u−4+(1/u)=0  u^2 −4u+1=0  u=2±(√3)  t^x^2  =2±(√3)  x^2 =log_(2+(√3))  (2±(√3))  if u=2+(√3),  x^2 =1  x=±1  if u=2−(√3),  x^2 =log_(2+(√3))  (1/(2+(√3)))=−1 (∵2−(√3)=(1/(2+(√3))))  x=±i  ∴x=±1 or ±i ■