Question

|x-2| > square root (x)

Answer 1

x = 0

Explanation:

I. When x cant be negative value and

II. Find possible positive value

If x = 0

|0-2| >
`√(0)`

|-2| > 0

2 > 0 true

If x = 1

|1 - 2| >
`√(1)`

| -1 | > 1

1 > 1 false

More question or something to discuss, leave comment below

Answer 2

Explanation:

x ≥ 0 (because the square root)

`|x-2| > √(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)\\\\`

`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 \\`

Sol= [0, 1] ∪ ]4,+∞) ***** corrected