$$\eqalign{ & {x^2} - x + 1 = 0 \cr & \Rightarrow \,\,x = \frac{{1 \pm \sqrt {1 - 4} }}{2} \cr & x = \frac{{1 \pm \sqrt 3 \,i}}{2} \cr & \alpha = \frac{1}{2} + i\frac{{\sqrt 3 }}{2} = - {\omega ^2}\,\,\,\,\,\,\,\,\beta = \frac{1}{2} - \frac{{i\sqrt 3 }}{2} = - \omega \cr & = {\alpha ^{2009}} + {\beta ^{2009}} \cr & = {\left( { - {\omega ^2}} \right)^{2009}} + {\left( { - \omega } \right)^{2009}} \cr & \, = - {\omega ^2} - \omega = 1 \cr} $$

Put $$x = - 2;$$   then $${\left( { - 2} \right)^3} + 14{\left( { - 2} \right)^2} - 84\left( { - 2} \right) - 216 = 0$$
∴ the equation is $$\left( {x + 2} \right)\left( {{x^2} + 12x - 108} \right) = 0\,\,{\text{or, }}\left( {x + 2} \right)\left( {x - 6} \right)\left( {x + 18} \right) = 0.$$
Roots are $$- 2, 6, - 18.$$

$$\eqalign{ & \left| {\left| {x - 1} \right| - 1} \right| \leqslant 1 \cr & \Rightarrow \,\,0 \leqslant \left| {x - 1} \right| \leqslant 2 \cr & \Rightarrow \,\,\left| {x - 1} \right| \leqslant 2 \cr & \Rightarrow \,\, - 1 \leqslant x \leqslant 3. \cr} $$

Given roots are real and distinct, then $$a^2 - 4b > 0$$
$$ \Rightarrow b < \frac{{{a^2}}}{4}$$
Again $$\alpha$$ and $$\beta$$ differ by a quantity less than $$c\left( {c > 0} \right)$$
$$\eqalign{ & \Rightarrow \left| {\alpha - \beta } \right| < c{\text{ or }}{\left( {\alpha - \beta } \right)^2} < {c^2} \cr & \Rightarrow {\left( {\alpha + \beta } \right)^2} - 4\alpha \beta < {c^2}{\text{ or }}{a^2} - 4b < {c^2}\,\,{\text{or }}\frac{{{a^2} - {c^2}}}{4} < b \cr & \Rightarrow \frac{{{a^2} - {c^2}}}{4} < b < \frac{{{a^2}}}{4}{\text{by}}\left( 1 \right){\text{and}}\left( 2 \right). \cr} $$

Solving the equations, $${x^3} + {x^2} + x - 14 = 0$$
$$\eqalign{ & \Rightarrow \,\,\left( {x - 2} \right)\left( {{x^2} + 3x + 7} \right) = 0 \cr & \therefore \,\,x = 2,\frac{{ - 3 \pm \sqrt {9 - 28} }}{2}. \cr} $$

$$\eqalign{ & \frac{{\left| {x + 1} \right|}}{{\left| x \right|}} + \left| {x + 1} \right| = \frac{{\left| {x + 1} \right|^2}}{{\left| x \right|}} \cr & \Rightarrow \,\,\left| {x + 1} \right|\left\{ {\frac{1}{{\left| x \right|}} + 1 - \frac{{\left| {x + 1} \right|}}{{\left| x \right|}}} \right\} = 0 \cr & \therefore \,\,\left| {x + 1} \right| = 0\,\,\,{\text{or, }}1 + \left| x \right| - \left| {x + 1} \right| = 0. \cr & \left| {x + 1} \right| = 0 \cr & \Rightarrow \,\,x = - 1. \cr & {\text{If }}x < - 1,1 + \left| x \right| - \left| {x + 1} \right| = 0 \cr & \Rightarrow \,\,1 - x + x + 1 = 0 \cr & \Rightarrow \,\,2 = 0\left( {{\text{absurd}}} \right). \cr & {\text{If }} - 1 \leqslant x < 0,1 + \left| x \right| - \left| {x + 1} \right| = 0 \cr & \Rightarrow \,\,1 - x - \left( {x + 1} \right) = 0 \cr & \Rightarrow \,\,x = 0\left( {{\text{not possible}}} \right). \cr & {\text{If }}x \geqslant 0,1 + x - \left( {x + 1} \right) = 0 \cr & \Rightarrow \,\,0 = 0 \cr} $$
⇒ $$x$$ can have any value in the interval.
$$\therefore \,\,x = - 1,x > 0\left( {\because \,\,x \ne 0} \right).$$

Since $$\,2 - \sqrt 3 $$   is a root of the quadratic equation
$${x^2} + px + q = 0$$
$$\therefore \,\,2 + \sqrt 3 \,$$   is the root of unity
⇒ Sum of roots = 4, Product of roots = 1
$$\eqalign{ & \Rightarrow \,p = - 4,q = 1 \cr & \Rightarrow \,{p^2} - 4q - 12 \cr & = 16 - 4 - 12 \cr & = 0 \cr} $$

$$\eqalign{ & {2^{\frac{x}{2}}} + {3^{\frac{x}{2}}} = {\left( {\sqrt {13} } \right)^{\frac{x}{2}}} \cr & \Rightarrow \,\,{\left( {\frac{2}{{\sqrt {13} }}} \right)^{\frac{x}{2}}} + {\left( {\frac{3}{{\sqrt {13} }}} \right)^{\frac{x}{2}}} = 1 \cr} $$
which is of the form $${\cos ^{\frac{x}{2}}}\alpha + {\sin ^{\frac{x}{2}}}\alpha = 1.$$
$$\therefore \,\,\frac{x}{2} = 2.$$

First of all for $$\log \left( {x - 1} \right)$$   to be defined, $$x - 1 > 0$$
$$\eqalign{ & \Rightarrow \,x > 1\,\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{Now, }}{\log _{0.3}}\left( {x - 1} \right) < {\log _{0.09}}\left( {x - 1} \right) \cr & \Rightarrow \,\,{\log _{0.3}}\left( {x - 1} \right) < {\log _{\left( {0.3} \right)}}^2\left( {x - 1} \right) \cr & \Rightarrow \,\,{\log _{0.3}}\left( {x - 1} \right) < \frac{1}{2}{\log _{0.3}}\left( {x - 1} \right) \cr & \Rightarrow \,\,2{\log _{0.3}}\left( {x - 1} \right) < {\log _{0.3}}\left( {x - 1} \right) \cr & \Rightarrow \,\,{\log _{0.3}}{\left( {x - 1} \right)^2} < {\log _{0.3}}\left( {x - 1} \right) \cr & \Rightarrow \,\,{\left( {x - 1} \right)^2} > \left( {x - 1} \right)\,\,\,\,\,\,\,\,{\bf{NOTE}}\,\,{\bf{THIS}}\,\,{\bf{STEP}} \cr & \,\,\,\,\,\,\left[ {{\text{The inequality is reversed since base lies between 0 and 1}}} \right] \cr & \Rightarrow \,{\left( {x - 1} \right)^2} - \left( {x - 1} \right) > 0 \cr & \Rightarrow \,\left( {x - 1} \right)\left( {x - 2} \right) > 0\,\,\,\,\,\,\,.....\left( 2 \right) \cr & \,\,\,\,\,{\text{Combining}}\left( 1 \right){\text{and}}\left( 2 \right){\text{we get}}\,\,x > 2 \cr & \therefore \,\,x \in \left( {2,\infty } \right) \cr} $$

By trial, $${3^{\frac{x}{2}}} + {2^x} \leqslant 25\,\,{\text{for }}x = 1,2,3,4.\,\,{\text{But }}{3^{\frac{x}{2}}} + {2^x} > 25\,\,{\text{for }}x > 4.$$