Let the second root be $$\alpha .$$

Then $$\alpha + \left( {1 - p} \right) = - p \Rightarrow \alpha = - 1$$

Also $$\alpha .\left( {1 - p} \right) = 1 - p$$

$$ \Rightarrow \left( {\alpha - 1} \right)\left( {1 - p} \right) = 0$$

$$ \Rightarrow p = 1$$ [as $$\alpha = - 1$$]

$$\therefore$$ Roots are $$\alpha = - 1$$ and $$p-1=0$$

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 $$=4$$