Let x−[x]=t
∴−3t2+2t+a2=0
As x is not an integer, t=0 ∴a=0
For a root of x to exist at least one of the roots of t should be between 0 and 1.
Roots =−6−2±4−(4)(−3)a2
=−6−2±4+12a2
=−3−1±1+3a2=31∓1+3a2
As we observe, one root is surely less than zero,
i.e., 31−1+3a2
∴ For a solution to exist,
31+1+3a2<1
∴1+1+3a2<3
∴1+3a2<2
∴1+3a2<4
∴3a2<3
∴a2<1
∴a∈(−1,0)∪(0,1). [∵a=0]