both roots are less than $$5,$$

then $$(i)$$ Discriminant $$ \ge 0$$

$$\left( {ii} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p\left( 5 \right) > 0$$

$$\left( {iii} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{Sum\,\,of\,\,roots} \over 2} < 5$$

Hence $$\left( i \right)\,\,\,\,\,\,4{k^2} - 4\left( {{k^2} + k - 5} \right) \ge 0$$

$$4{k^2} - 4{k^2} - 4k + 20 \ge 0$$

$$4k \le 20 \Rightarrow k \le 5$$

$$\left( {ii} \right)\,\,\,\,\,f\left( 5 \right) > 0;25 - 10k + {k^2} + k - 5 > 0$$

or $${k^2} - 9k + 20 > 0$$

or $$k\left( {k - 4} \right) - 5\left( {k - 4} \right) > 0$$

or $$\left( {k - 5} \right)\left( {k - 4} \right) > 0$$

$$ \Rightarrow k \in \left( { - \infty ,4} \right) \cup \left( { - \infty ,5} \right)$$

$$\left( {iii} \right)\,\,\,\,\,\,{{Sum\,\,of\,\,roots} \over 2}$$

$$ = - {b \over {2a}} = {{2k} \over 2} < 5$$

The intersection of $$(i)$$, $$(ii)$$ & $$(iii)$$ gives

$$k \in \left( { - \infty ,4} \right).$$

$$\angle $$R = 90^o $$ \therefore $$ $$\angle $$P + $$\angle $$Q = 90^o

$$ \Rightarrow $$ $${P \over 2} + {Q \over 2} = {{90} \over 2} = 45$$^o

$$\tan \left( {{P \over 2}} \right),\tan \left( {{Q \over 2}} \right)$$ are the roots of $$a{x^2} + bx + c = 0$$

$$ \therefore $$ $$\tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right) = - {b \over a},\,\,$$

and $$\tan \left( {{P \over 2}} \right).\tan \left( {{Q \over 2}} \right) = {c \over a}$$

$${{\tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right)} \over {1 - \tan \left( {{P \over 2}} \right)\tan \left( {{Q \over 2}} \right)}}$$

$$ = \tan \left( {{P \over 2} + {Q \over 2}} \right) $$= tan 45^o = 1

$$ \Rightarrow {{ - {b \over a}} \over {1 - {c \over a}}} = 1$$

$$ \Rightarrow - {b \over a} = {a \over a} - {c \over a}$$

$$ \Rightarrow - b = a - c$$ or $$c = a + b.$$

$$4$$ is a root of $${x^2} + px + 12 = 0$$

$$ \Rightarrow 16 + 4p + 12 = 0$$

$$ \Rightarrow p = - 7$$

Now, the equation $${x^2} + px + q = 0$$

has equal roots.

$$\therefore$$ $${p^2} - 4q = 0$$ $$ \Rightarrow q = {{{p^2}} \over 4} = {{49} \over 4}$$

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 $$p-1=0$$