The equation `(x)^(2)=[x]^(2)+2x` where `[x]` and `(x)` are the integers just less than or equal to `x` and just greater than or equal to `x` respectively, then number of values of `x` satisfying the given equation

Explanation

Case I If `x epsilonI` then
`x=[x]=(x)`
The given equation reduces to
`x^(2)=x^(2)+2x`
`implies2x=0` or `x=0`….i
Case II If `x!inI`, then `(x)=[x]+1`
The given equation reduces to
`([x]+1)^(2)=[x]^(2)+2x`
`implies1=2(x-[x])` or `{x}=1/2`
`:.x=[x]+1/2=n+1/2,n epsilonI`............ii ,brgt Hence the solution of the original equation is `x=0, n+1/2, n epsilon I`.