We have, √x+3−4√x−1+√x+8−6√x−1=1 √(√x−1)2−2×2√x−1+4+√(√x−1)2−2×3√x−1+9=1 ⇒√(√x−1−2)2+√(√x−1−3)2=1 ⇒|√x−1−2|+|√x−1−3|=1 ⇒|√x−1−2|+|√x−1−3| =(√x−1−2)−(√x−1−3) We know that, If |x−a|+|x−b|=(x−a)−(x−b) then, (x−a)(x−b)<0 ∴ (√x−1−2)(√x−1−3)<0 ⇒2<√x−1<3⇒5<x<10 ∴ Equation have infinite many solutions.