Solve the equation `sqrt(x)=x-2`

Explanation

We have `sqrt(x)=x-2`
On squaring both sides we obtain
`x=(x-2)^(2)`
`impliesx^(2)-5x+4=0implies(x-1)(x-4)=0`
`:.x_(1)=1` and `x_(2)=4`
Hence `x_(1)=4` satisfies the original equation, but `x_(2)=1` does not satisfy the original equation.
`:.x_(2)=1` is the extraneous root.