If a ∈ R and the equation - 3 ( x - [ x ] ) ^ 2 + 2 ( x - [ x ] ) +...

Let $x - [ x ] = t$

$∴ -3 {t}^{2} + 2t + {a}^{2} = 0$

As $x$ is not an integer, $t \ne 0 ∴ a \ne 0$

For a root of $x$ to exist at least one of the roots of $t$ should be between $0 and 1$.

Roots $=\dfrac{- 2 \pm \sqrt{4 - \left(4\right)\left(- 3\right){a}^{2}}}{- 6}$

$= \dfrac{- 2 \pm \sqrt{4 + 12{a}^{2}}}{- 6}$

$= \dfrac{- 1 \pm \sqrt{1 + 3{a}^{2}}}{- 3} = \dfrac{1 ∓ \sqrt{1 + 3{a}^{2}}}{3}$

As we observe, one root is surely less than zero,

i.e., $\dfrac{1 - \sqrt{1 + 3{a}^{2}}}{3}$

$∴$ For a solution to exist,

$\dfrac{1 + \sqrt{1 + 3{a}^{2}}}{3} < 1$

$∴ 1 + \sqrt{1 + 3{a}^{2}} < 3$

$∴ \sqrt{1 + 3{a}^{2}} < 2$

$∴ 1 + 3{a}^{2} < 4$

$∴ 3{a}^{2} < 3$

$∴ {a}^{2} < 1$

$∴ a \in \left(- 1, 0\right) \cup \left(0, 1\right). \left[∵ a \ne 0\right]