JEE Main 8 Apr 2019 Paper 1 Solved Paper

Case 1: When √x≥2 then |√x−2|=√x−2
∴ The given equation becomes,
‌(√x−2)+√x(√x−4)+2=0
⇒(√x−2)+x−4√x+2=0
⇒x−3√x=0
⇒√x(√x−3)=0
∴√x=0 or 3
√x=0 is not possible as √x≥2.
So, √x=3
or x=9
Case 2 : When √x<2
then |√x−2|=−(√x−2)=2−√x
∴ The given equation becomes,
(2−√x)+√x(√x−4)+2=0
⇒2−√x+x−4√x+2=0
⇒x−5√x+4=0
⇒x−4√x−√x+4=0
⇒√x(√x−4)−(√x−4)=0
⇒(√x−4)(√x−1)=0
∴√x=4 or 1
√x=4 is not possible as √x<2
∴√x=1
or x=1
So, Sum of all solutions =9+1=10