Given that roots of the equation
$$\eqalign{ & b{x^2} + cx + a = 0\,{\text{are}}\,{\text{imaginary}} \cr & \therefore \,\,\,\,\,{c^2} - 4ab < 0\,\,\,\,.....\left( {\text{i}} \right) \cr & {\text{Let}}\,y = 3{b^2}{x^2} + 6bcx + 2{c^2} \cr & \Rightarrow \,\,3{b^2}{x^2} + 6bcx + 2{c^2} - y = 0 \cr & {\text{As}}\,x\,{\text{is}}\,{\text{real}},D \geqslant 0 \cr & \Rightarrow \,\,\,36{b^2}{c^2} - 12{b^2}\left( {2{c^2} - y} \right) \geqslant 0 \cr & \Rightarrow \,\,\,12{b^2}\left( {3{c^2} - 2{c^2} + y} \right) \geqslant 0 \cr & \Rightarrow \,\,\,{c^2} + y \geqslant 0 \cr & \Rightarrow \,\,y \geqslant - {c^2} \cr & {\text{But}}\,{\text{from}}\,{\text{eqn}}{\text{.}}\,\left( {\text{i}} \right),\,{c^2} < 4ab\,\,{\text{or}}\,\, - {c^2} > - 4ab \cr & \therefore \,\,\,{\text{we}}\,{\text{get}}\,y \geqslant - {c^2} > - 4ab \cr & \Rightarrow \,\,y > - 4ab \cr} $$

Take the equation $$2a{x^3} + 3b{x^2} + 6cx = 0.$$     Clearly, $$x = 0, 1$$   are two roots. Observe that $$a{x^2} + bx + c = 0$$    is the derived equation. So, one root lies between 0 and 1.

$$\eqalign{ & f\left( x \right) = 2{x^3} + 3x + k \cr & f'\left( x \right) = 6{x^2} + 3 > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\forall \,x \in R\,\,\left( {\because \,\,{x^2} > 0} \right) \cr} $$
$$ \Rightarrow \,\,f\left( x \right)$$  is strictly increasing function
$$ \Rightarrow \,\,f\left( x \right) = 0$$   as only one real root, so two roots are not possible.

$${5^x} + {5^{ - x}} \geqslant 2\left( {\because \,\,{\text{AM }} \geqslant {\text{ GM}}} \right);\,{\text{but }}\sin {e^x} \leqslant 1.$$

For, $$x \geqslant - 2,{x^2} - x - 2 + x > 0$$
$$\eqalign{ & \Rightarrow {x^2} > 2 \cr & \Rightarrow x \in \left( { - \infty , - \sqrt 2 } \right) \cup \left( {\sqrt 2 ,\infty } \right) \cr & \Rightarrow x \in \left[ { - 2, - \sqrt 2 } \right) \cup \left( {\sqrt 2 ,\infty } \right)\,\,{\text{For, }}x < - 2 \cr & {x^2} + x + 2 + x > 0{\text{ or }}{x^2} + 2x + 2 > 0 \cr} $$
which is true for all $$x.$$
Hence, $$x \in \left( { - \infty , - \sqrt 2 } \right) \cup \left( {\sqrt 2 ,\infty } \right)$$

Case 1 : $$x \geqslant 0$$
∴ the equation becomes $$x^2 + 5x + 4 =0$$    or $$x = - 1, - 4$$   but $$x \geqslant 0$$
∴ both values, non admissible :
Case 2 : $$x \leqslant 0$$
The equation becomes $$x^2 - 5x + 4 = 0$$    or $$x = 1, 4$$   both values are non admissible
∴ No real roots.
Alternatively, since $${x^2} \geqslant 0;\left| x \right| \geqslant 0$$
$$\eqalign{ & \therefore {x^2} + \left| x \right| + 4 > 0{\text{ for all }}x \in {\bf{R}} \cr & \therefore {x^2} + \left| x \right| + 4 \ne 0{\text{ for any }}x \in {\bf{R}} \cr} $$

Consider both equations
$$\eqalign{ & p{x^2} + 2qx + r = 0\,\,\,\,\,.....\left( {\text{i}} \right) \cr & {\text{and, }}q{x^2} - 2\sqrt {pr} .x + q = 0\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Since, both the equations are quadratic and have real roots, therefore from equation (i), we have
$$\eqalign{ & \therefore 4{q^2} - 4pr \geqslant 0\left( {{\text{using discriminant}}} \right) \cr & \Rightarrow {q^2} \geqslant pr\,\,\,\,.....\left( {{\text{iii}}} \right) \cr} $$
and from second equation $$4pr - 4{q^2} \geqslant 0$$
$$ \Rightarrow pr \geqslant {q^2}\,\,\,.....\left( {{\text{iv}}} \right)$$
From equations (iii) and (iv) we get, $$q^2 = pr.$$

As $$a\left( {b - c} \right) + b\left( {c - a} \right) + c\left( {a - b} \right) = 0,x = 1$$         is a root of the corresponding equation. The other root of the equation
$$ = \frac{{c\left( {a - b} \right)}}{{a\left( {b - c} \right)}} = 1$$    because $$a, b, c$$  in H.P. implies $$\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b},\,{\text{i}}{\text{.e}}{\text{., }}\frac{{a - b}}{a} = \frac{{b - c}}{c}$$
∴ $$x = 1, 1$$   are the roots of the corresponding equation. So, $${\left( {x - 1} \right)^2}$$  is a factor.

$${\left( {\frac{{\sqrt 2 }}{{\sqrt {5 + 2\sqrt 2 } }}} \right)^x} + \left( {\frac{{\sqrt 2 + 1}}{{\sqrt {5 + 2\sqrt 2 } }}} \right) = 1$$        which is of the form
$${\cos ^x}\alpha + {\sin ^x}\alpha = 1;\,\,\,\therefore x = 2.$$

Let $$a = \sqrt {{x^2} + x} \,\,{\text{and}}\,\,b = \frac{{{{\tan }^2}\alpha }}{{\sqrt {{x^2} + x} }}$$
then using A.M. $$ \geqslant $$ G.M., we get $$\frac{{a + b}}{2} \geqslant \sqrt {ab} $$
$$\eqalign{ & \Rightarrow \,\,a + b \geqslant 2\sqrt {ab} \cr & \Rightarrow \,\,\sqrt {{x^2} + x} + \frac{{{{\tan }^2}\alpha }}{{\sqrt {{x^2} + x} }} \geqslant 2\sqrt {{{\tan }^2}\alpha } = 2\tan \alpha \cr & \left[ {\because \,\,\alpha \in \left( {0,\frac{\pi }{2}} \right)} \right]. \cr} $$