Product of real quadratic roots If product of roots '$p$' of equation $x^2 – 2ax + 8 – a = 0$ lies between its roots, then maximum
integral value of '$p$' is _____
My approach is as follow
$f\left( x \right) = {x^2} - 2ax + 8 - a$
${D^2} = 4{a^2} - 4\left( {8 - a} \right) \ge 0 \Rightarrow {a^2} + a - 8 \ge 0$
${a^2} + a - 8 \ge 0 \Rightarrow a \in \left( { - \infty ,\frac{{ - 1 - \sqrt {33} }}{2}} \right) \cup \left( {\frac{{ - 1 + \sqrt {33} }}{2},\infty } \right)$
$\alpha  + \beta  = 2a$ & $\alpha \beta  = 8 - a$
$\alpha \beta  \in \left( {\alpha ,\beta } \right)$ where $\beta  > \alpha $
How do we proceed from hered

The roots are
$$a\pm \sqrt{a^2-(8-a)}$$
and the product of the roots is $8-a$, so defining $p=8-a$, we wish to maximize $p$ (among integers) subject to constraints
$$ 8-p-\sqrt{(8-p)^2-p}\leq p\leq 8-p+ \sqrt{(8-p)^2-p}.$$
A quick sketch shows optimally the upper constraint must bind so
$$p= 8-p+ \sqrt{(8-p)^2-p}\implies p=5.$$

