Answer:
Step-by-step explanation:
x ≥ 0 (because the square root)
[tex]|x-2| > \sqrt{x}\\\\1)\ if \ x-2 >0 \ (\ or\ x > 2):\\|x-2|=x-2\\\\x-2 > \sqrt(x) \Longrightarrow\ (x-2)^2 > x\\\Longrightarrow\ x^2-4x+4 > x\\\Longrightarrow\ x^2-5x+4 > 0\\\Longrightarrow\ (x-1)(x-4) > 0\\\Longrightarrow\ x<1\ or\ x>4 \Longrightarrow\ x >4 \ (since\ x>2)\\\\[/tex]
[tex]2)\ if\ x-2 <0 \ (\ or\ x < 2):\\|x-2|=-(x-2)=-x+2\\\\-x+2 > \sqrt(x) \Longrightarrow\ (-x+2)^2 > x\\\Longrightarrow\ x^2-4x+4 > x\\\Longrightarrow\ x^2-5x+4 > 0\\\Longrightarrow\ (x-1)(x-4) > 0\\\Longrightarrow\ x<1\ or\ x>4 \Longrightarrow\ x\geq 0\ and \ x\leq 1 \\[/tex]
Sol= [0, 1] ∪ ]4,+∞) ***** corrected
