Question Number 149876 by liberty last updated on 08/Aug/21

Answered by MJS_new last updated on 08/Aug/21

y≥0 y=(√(x+2(√(x−1))))+(√(x−2(√(x−1)))) squaring y^2 =(x+2(√(x−1)))+2(√((x+2(√(x−1)))(x−2(√(x−1)))))+(x−2(√(x−1))) y^2 =2x+2(√(x^2 −4(x−1))) y^2 =2x+2(√((x−2)^2 )) y^2 =2x+2∣x−2∣ x≤2 ⇒ ∣x−2∣=2−x y^2 =2x+2(2−x) y^2 =4 y>0 ⇒ y=2

Answered by john_santu last updated on 08/Aug/21

method 2 let 2(√(x−1)) = u ; x−1=(1/4)u^2 ⇒x=((u^2 +4)/4) ; 1≤((u^2 +4)/4)≤2 ⇒4≤u^2 +4≤8 ; 0≤u^2 ≤4 ⇒0≤u≤2 the equality become to (√(((u^2 +4)/4)+u)) +(√(((u^2 +4)/4)−u)) = (1/2) [(√(u^2 +4u+4)) +(√(u^2 −4u+4)) ]= (1/2)[ ∣u+2∣ +∣u−2∣ ] = (1/2)[ u+2−(u−2)] =(1/2)×4=2