Let two numbers be a and b then $${{a + b} \over 2} = 9$$

and $$\sqrt {ab} = 4$$

$$\therefore$$ Equation with roots $$a$$ and $$b$$ is

$${x^2} - \left( {a + b} \right)x + ab = 0$$

$$ \Rightarrow {x^2} - 18x + 16 = 0$$

$$y = x + {1 \over x}$$ or $${{dy} \over {dx}} = 1 - {1 \over {{x^2}}}$$

For max. or min, $$1 - {1 \over {{x^2}}} = 0 \Rightarrow x = \pm 1$$

$${{{d^2}y} \over {d{x^2}}} = {2 \over {{x^3}}} \Rightarrow {\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{x = 2}} = 2$$ ($$+ve$$ minima)

$$\therefore$$ $$x=1$$

$${x^2} - 3\left| x \right| + 2 = 0$$

$$ \Rightarrow {\left| x \right|^2} - 3\left| x \right| + 2 = 0$$

$$\left( {\left| x \right| - 2} \right)\left( {\left| x \right| - 1} \right) = 0$$

$$\left| x \right| = 1,2$$ or $$x = \pm 1, \pm 2$$

$$\therefore$$ No. of solution $$=4$$

Let the roots of given equation be $$\alpha $$ and $$2$$$$\alpha $$ then

$$\alpha + 2\alpha = 3\alpha = {{1 - 3a} \over {{a^2} - 5a + 3}}$$

and $$\alpha .2\alpha = 2{\alpha ^2} = {2 \over {{a^2} - 5a + 3}}$$

$$ \Rightarrow \alpha = {{1 - 3a} \over {3\left( {{a^2} - 5a + 3} \right)}}$$

$$\therefore$$ $$2\left[ {{1 \over 9}{{{{\left( {1 - 3a} \right)}^2}} \over {{{\left( {{a^2} - 5a + 3} \right)}^2}}}} \right]$$

$$ = {2 \over {{a^2} - 5a + 3}}$$

$${{{{\left( {1 - 3a} \right)}^2}} \over {\left( {{a^2} - 5a + 3} \right)}} = 9$$

or $$9{a^2} - 6a + 1$$

$$ = 9{a^2} - 45a + 27$$

or $$39a = 26$$ or $$a = {2 \over 3}$$