Consider $$ - 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0\,$$
$$\eqalign{ & \Rightarrow \,3{\left\{ x \right\}^2} - 2\left\{ x \right\} - {a^2} = 0\,\,\,\,\,\,\,\left( {\because \,x - \left[ x \right] = \left\{ x \right\}} \right) \cr & \Rightarrow \,\,3\left( {{{\left\{ x \right\}}^2} - \frac{2}{3}\left\{ x \right\}} \right) = {a^2},a \ne 0 \cr & \Rightarrow \,\,{a^2} = 3\left\{ x \right\}\left( {\left\{ x \right\} - \frac{2}{3}} \right) \cr} $$

$${\text{Now,}}\,\,\left\{ x \right\} \in \left( {0,1} \right)\,\,{\text{and }}\frac{{ - 2}}{3} \leqslant {a^2} < 1\,\,\,\,\left( {{\text{by graph}}} \right)$$
Since, $$x$$ is not an integer
$$\eqalign{ & \therefore \,\,a \in \left( { - 1,1} \right) - \left\{ 0 \right\} \cr & \Rightarrow \,\,a \in \left( { - 1,0} \right) \cup \left( {0,1} \right) \cr} $$

From the first method,
$$\eqalign{ & q = \alpha \beta ,r = {\alpha ^4} + {\beta ^4}\,\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{Product of the roots of the equation}} \cr & {x^2} - 4qx + \left( {2{q^2} - r} \right) = 0 \cr & = 2{q^2} - r = 2{\alpha ^2}{\beta ^2} - {\alpha ^4} - {\beta ^4} \cr & = - {\left( {{\alpha ^2} - {\beta ^2}} \right)^2}\,\,\,\,\,\,\left[ {{\text{From}}\left( 1 \right)} \right] \cr & = - \left( {{\text{positive quantity}}} \right) \cr & = - ve{\text{ quantity}} \cr & \Rightarrow \,\,{\text{one root is positive and other is negative}}{\text{.}} \cr} $$

$$\eqalign{ & \alpha < - 1.\,{\text{Let }}\alpha = - 1 - p \cr & \beta > 1.\,{\text{Let }}\beta = 1 + q,p > 0,q > 0 \cr & {\text{Now, }}1 + \frac{c}{a} + \left| {\frac{b}{a}} \right| = 1 + \alpha \beta + \left| { - \alpha - \beta } \right| \cr & = 1 + \left( {1 + q} \right)\left( { - 1 - p} \right) + \left| { - 1 - p + 1 + q} \right| \cr & = 1 - \left( {1 + p + q + pq} \right) + \left| {q - p} \right| \cr} $$
\[ = \left\{ \begin{gathered} - p - q - pq + q - p = - 2p - pq < 0\,\,\,{\text{if, }}q > p \hfill \\ - p - q - pq + p - q = - 2q - pq < 0\,\,\,{\text{if,}}\,\,q < p \hfill \\ \end{gathered} \right.\]
$$\therefore 1 + \frac{c}{a} + \left| {\frac{b}{a}} \right| < 0$$

Let $$f\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx - 1.$$
$${\left( {x - 1} \right)^3}$$  is a factor
$$ \Rightarrow \,\,f\left( 1 \right) = 0,f'\left( 1 \right) = 0,f''\left( 1 \right) = 0.$$
This given $$a, b, c.$$

$$\eqalign{ & \left( {1 + \lambda } \right){x^2} - \left( {6 + 4\lambda } \right)x + 8 + 3\lambda = 0 \cr & \therefore \,\,D = {\left( {6 + 4\lambda } \right)^2} - 4\left( {1 + \lambda } \right)\left( {8 + 3\lambda } \right) = {\lambda ^2} + \lambda + 1 > 0. \cr} $$

$${\text{Use }}f\left( { - \frac{1}{2}} \right) = 5\,\,{\text{and}}\,f'\left( {\frac{1}{3}} \right) = 0.$$

As $$\alpha ,\beta $$  are the roots of $${x^2} - px + r = 0$$
$$\eqalign{ & \therefore \alpha + \beta = p\,\,\,\,\,\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{and }}\alpha \beta = r\,\,\,\,\,\,\,.....\left( 2 \right) \cr} $$
Also $$\frac{\alpha }{2},2\beta $$   are the roots of $${x^2} - qx + r = 0$$
$$\therefore \,\,\frac{\alpha }{2} + 2\beta = q\,\,{\text{or }}\alpha + 4\beta = 2q\,\,\,\,\,\,\,.....\left( 3 \right)$$
Solving (1) and (3) for $$\alpha $$ and $$\beta $$ , we get
$$\beta = \frac{1}{3}\left( {2q - p} \right)\,\,{\text{and }}\alpha = \frac{2}{3}\left( {2q - q} \right)$$
Substituting values of $$\alpha $$ and $$\beta $$ , in equation (2),
we get $$\frac{2}{9}\left( {2p - q} \right)\left( {2q - p} \right) = r.$$

If $$\alpha ,\beta $$  be the roots of the equation then $$\alpha + \beta = a - 2,\alpha \beta = - \frac{{a + 1}}{2}$$
Sum of square of roots
$$\eqalign{ & S = {\alpha ^2} + {\beta ^2} = \left( {\alpha + {\beta ^2}} \right) - 2\alpha \beta \cr & = {\left( {a - 2} \right)^2} + \left( {a + 1} \right) = {a^2} - 3a + 5 \cr & S = {a^2} - 3a + \frac{9}{4} + \frac{{11}}{4} \cr & S = {\left( {a - \frac{3}{2}} \right)^2} + \frac{{11}}{4} \cr} $$
Clearly $$S$$ is least when
$$\eqalign{ & a - \frac{3}{2} = 0 \cr & \Rightarrow a = \frac{3}{2} \cr} $$

For real roots $${q^2} - 4pr \geqslant 0$$
$$\eqalign{ & \Rightarrow \,\,{\left( {\frac{{p + r}}{2}} \right)^2} - 4pr \geqslant 0\,\,\,\,\,\,\,\,\,\left( {\because \,p,q,r\,\,{\text{are in A}}{\text{.P}}{\text{.}}} \right) \cr & \Rightarrow \,\,{p^2} + {r^2} - 14pr \geqslant 0 \cr & \Rightarrow \,\,\frac{{{p^2}}}{{{r^2}}} - 14\frac{p}{r} + 1 \geqslant 0 \cr & \Rightarrow \,\,{\left( {\frac{p}{r} - 7} \right)^2} - 48 \geqslant 0 \cr & \Rightarrow \,\,\left| {\frac{p}{r} - 7} \right| \geqslant 4\sqrt 3 \cr} $$

Let two numbers be $$a$$ and $$b$$ then $$\frac{{a + b}}{2} = 9\,\,{\text{and }}\sqrt {ab} = 4$$
∴ Equation with roots $$a$$ and $$b$$ is
$$\eqalign{ & {x^2} - \left( {a + b} \right)x + ab = 0 \cr & \Rightarrow \,\,{x^2} - 18x + 16 = 0 \cr} $$