Since the roots are imaginary
∴ $$D < 0$$  and roots occur as conjugate pair, i.e., $$\beta = \overline \alpha $$
$$\eqalign{ & \therefore \left| \beta \right| = \left| {\overline \alpha } \right| = \left| \alpha \right| \cr & {\text{Also, let }}\alpha = \frac{{ - b + i\sqrt {4ac - {b^2}} }}{{2a}} \cr & \therefore \left| \alpha \right| = \sqrt {\frac{{{b^2}}}{{4{a^2}}} + \frac{{4ac - {b^2}}}{{4{a^2}}}} = \sqrt {\frac{c}{a}} \cr & \left| \alpha \right| > 1\left( {\because c > a} \right) \cr & \therefore \left| \alpha \right| = \left| \beta \right| > 1 \cr} $$

Here $${b^2} - 24a \leqslant 0.\,{\text{Let }}3a + b = \lambda .$$
Then $${b^2} - 8\left( {\lambda - b} \right) \leqslant 0\,\,{\text{or, }}{b^2} + 8b - 8\lambda \leqslant 0.$$
Since $$b$$ is real, therefore, $$64 + 32\lambda \geqslant 0$$
$$\therefore \,\,\lambda \geqslant - 2.$$

$$\because \,\,a,b,c$$   are sides of a triangle and $$a \ne b \ne c$$
$$\therefore \,\,\left| {a - b} \right| < \left| c \right|$$
$$\eqalign{ & \Rightarrow \,\,{a^2} + {b^2} - 2ab < {c^2} \cr & {\text{Similarly, we get}} \cr & {b^2} + {c^2} - 2bc < {a^2};{c^2} + {a^2} - 2ca < {b^2} \cr & {\text{On adding, we get}} \cr & {a^2} + {b^2} + {c^2} < 2\left( {ab + bc + ca} \right) \cr & \Rightarrow \,\,\frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ca}} < 2\,\,\,\,\,\,\,.....\left( 1 \right) \cr} $$
$$\because $$ Roots of the given equation are real
$$\eqalign{ & \therefore \,\,{\left( {a + b + c} \right)^2} - 3\lambda \left( {ab + bc + ca} \right) \geqslant 0 \cr & \Rightarrow \,\,\frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ca}} \geqslant 3\lambda - 2\,\,\,\,\,\,\,\,\,\,.....\left( 2 \right) \cr & {\text{From}}\left( 1 \right){\text{and}}\left( 2 \right),{\text{we get}} \cr & {\text{3}}\lambda - 2 < 2 \cr & \Rightarrow \,\,\lambda < \frac{4}{3}. \cr} $$

$$\eqalign{ & \alpha ,\beta = \frac{{6 \pm \sqrt {36 + 8} }}{2} = 3 \pm \sqrt {11} \cr & \alpha = 3 + \sqrt {11} ,\beta = 3 - \sqrt {11} \cr & \therefore \,\,{a_n} = {\left( {3 + \sqrt {11} } \right)^n} - {\left( {3 - \sqrt {11} } \right)^n} \cr & \frac{{{a_{10}} - 2{a_8}}}{{2{a_9}}} \cr & = \frac{{{{\left( {3 + \sqrt {11} } \right)}^{10}} - {{\left( {3 - \sqrt {11} } \right)}^{10}} - 2{{\left( {3 + \sqrt {11} } \right)}^8} + 2{{\left( {3 - \sqrt {11} } \right)}^8}}}{{2\left[ {{{\left( {3 + \sqrt {11} } \right)}^9} - {{\left( {3 - \sqrt {11} } \right)}^9}} \right]}} \cr & = \frac{{{{\left( {3 + \sqrt {11} } \right)}^8}\left[ {{{\left( {3 + \sqrt {11} } \right)}^2} - 2} \right] + {{\left( {3 - \sqrt {11} } \right)}^8}\left[ {2 - {{\left( {3 - \sqrt {11} } \right)}^2}} \right]}}{{2\left[ {{{\left( {3 + \sqrt {11} } \right)}^9} - {{\left( {3 - \sqrt {11} } \right)}^9}} \right]}} \cr & = \frac{{{{\left( {3 + \sqrt {11} } \right)}^8}\left( {9 + 11 + 6\sqrt {11} - 2} \right) + {{\left( {3 - \sqrt {11} } \right)}^8}\left( {2 - 9 - 11 + 6\sqrt {11} } \right)}}{{2\left[ {{{\left( {3 + \sqrt {11} } \right)}^9} - {{\left( {3 - \sqrt {11} } \right)}^9}} \right]}} \cr & = \frac{{6{{\left( {3 + \sqrt {11} } \right)}^9} - 6{{\left( {3 - \sqrt {11} } \right)}^9}}}{{2\left[ {{{\left( {3 + \sqrt {11} } \right)}^9} - {{\left( {3 - \sqrt {11} } \right)}^9}} \right]}} = \frac{6}{2} = 3 \cr} $$

$$\eqalign{ & {\text{The given quadric equation is }}{x^2} + \left| x \right| - 6 = 0 \cr & {\text{Here, }}a = 1,{\text{ }}b = \pm 1{\text{ and }}c = - 6 \cr & {\text{As we know that }}D = {b^2} - 4ac \cr & {\text{Putting the value of }}a = 1,{\text{ }}b = \pm 1{\text{ and }}c = - 6 \cr & = {\left( { \pm 1} \right)^2} - 4 \times 1 \times - 6 \cr & = 1 + 24 \cr & = 25 \cr & {\text{Since, }}D \geqslant 0 \cr & {\text{Therefore, root of the given equation are real and distinct}}{\text{. }} \cr & {\text{Thus, sum of the roots be}} = 0. \cr} $$

$$\eqalign{ & {x^2} + px + q = 0 \cr & {\text{Let roots be }}\alpha {\text{ and }}{\alpha ^2} \cr & \Rightarrow \,\,\alpha + {\alpha ^2} = - p,\alpha {\alpha ^2} = q \cr & \Rightarrow \,\,\alpha = {q^{\frac{1}{3}}} \cr & \therefore \,\,{\left( q \right)^{\frac{1}{3}}} + {\left( {{q^{\frac{1}{3}}}} \right)^2} = - p \cr & {\text{Taking cube of both sides,}} \cr & {\text{we get}}\,\,\,q + {q^2} + 3q\left( {{q^{\frac{1}{3}}} + {q^{\frac{2}{3}}}} \right) = - {p^3} \cr & \Rightarrow \,\,q + {q^2} - 3pq = - {p^3} \cr & \Rightarrow \,\,{p^3} + {q^2} - q\left( {3p - 1} \right) = 0 \cr} $$

$$2{x^3} + 6{x^2} + x + 2 = 0$$     has roots $$\alpha ,\beta ,\gamma .$$
So, $$2{x^3} + {x^2} + 6x + 2 = 0$$     has roots $${\alpha ^{ - 1}},{\beta ^{ - 1}},{\gamma ^{ - 1}}$$   (writing co-efficients in reverse order).
$$\therefore \,\,{\alpha ^{ - 1}} + {\beta ^{ - 1}} + {\gamma ^{ - 1}} = - \frac{1}{2}.$$

Given equation is $$x^2 + px + q = 0$$
Sum of roots $$ = \tan {30^ \circ } + \tan {15^ \circ } = - p$$
Product of roots $$ = \tan {30^ \circ } . \tan {15^ \circ } = q$$
$$\eqalign{ & \tan {45^ \circ } = \frac{{\tan {{30}^ \circ } + \tan {{15}^ \circ }}}{{1 - \tan {{30}^ \circ }.\tan {{15}^ \circ }}} = \frac{{ - p}}{{1 - q}} = 1 \cr & \Rightarrow - p = 1 - q \cr & \Rightarrow q - p = 1 \cr & \therefore 2 + q - p = 3 \cr} $$

The sum of the co-efficients $$ = \left( {a + 2} \right) + \left( {a - 3} \right) - \left( {2a - 1} \right) = 0.\,{\text{So, }}x = 1$$         is a root.
∴ other root $$ = - \frac{{2a - 1}}{{a + 2}} = $$    rational for all $$a,a \ne - 2.$$

$$\eqalign{ & {\left( {\sqrt {\frac{\alpha }{\beta }} + \sqrt {\frac{\beta }{\alpha }} } \right)^2} = \frac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }} + 2 = \frac{{{{\left( {\alpha + \beta } \right)}^2}}}{{\alpha \beta }} \cr & = \frac{{{{\left( { - \frac{b}{a}} \right)}^2}}}{{\left( {\frac{b}{a}} \right)}} = \frac{b}{a} \cr & \therefore \sqrt {\frac{\alpha }{\beta }} + \sqrt {\frac{\beta }{\alpha }} = \sqrt {\frac{b}{a}} \,\,\,\,\left[ {\because \alpha ,\beta {\text{ are real}}} \right] \cr & \sqrt {\frac{\alpha }{\beta }} + \sqrt {\frac{\beta }{\alpha }} + \sqrt {\frac{b}{a}} = 2\sqrt {\frac{b}{a}} \cr} $$