The solution set of x which satisfies the equation `x^2 + (x+1)^2 = 25` where `(x)` is a least integer greater than or equal to x

Explanation

Case I If `x epsilonI` then `x=(x)=[x]`
Then `(x)^(2)+(x+1)^(2)=25` reduces to
`x^(2)+bar(x+1)^(2))=25implies2x^(2)+2x-24=0`
`impliesx^(2)+x-12=0implies(x+4)(x-3)=0`
`:.x=-4,3` ……..i
Case II If `x!inI`, then `(x)=[x]+1`
Then `(x)^(2)+(x+1)^(2)=25` reduces to
`{(x]+1}^(2)+{[x+1]+1}^(2)=25`
`implies{[x]+1}^(2)+{[x]+2}^(2)=25`
`implies2[x]^(2)+6[x]-20=0`
`=[x]^(2)+3[x]=10=0`
`implies{[x]+5}{[x]-2}=0`
`:.[x]=-5`and `[x]=2`
`impliesx epsilon[-5,-4)uu[2,3)`
`:' x !inI`
`:.x epsilon(-5,-4)uu(2,3)` ...........ii
On combining Eqs. i and ii we get
`x epsilonn (-5,-4]uu(2,3]`