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Q.
The value of P for which both the roots of the equation $4x^{2}-20\,px+(25 P^{2}+15p-66)=0$ are less than 2 lies in
Complex Numbers and Quadratic Equations
Solution:
Let $f (x) = 4x^{2}-20 Px +(25 P^{2}+15P-66)=0 \, \dots(i)$
As roots of (i) are real $\therefore D \ge\,0$
$\therefore 400P^{2}-16 (25 P^{2}+15P-66) \ge\,0$
$\therefore P \ge \frac{22}{5}$
Now, roots of (i) are less than 2
$\therefore f (2) >\,0$ and $\alpha+\beta<\,4$
$\therefore 16-40P+25 P^{2}+15P-66 >\, 0$ and $\frac{20}{4} P<\,4$
$\Rightarrow P^{2}-P-2 >\,0$ and $P<\, \frac{4}{5}$
$\therefore P>\,2$ or $P<\,-1$ and $P<\,\frac{4}{5}$
i.e. $P >\, 2$, or $P <\,-1$ and $P<\,\frac{4}{5}$
$\therefore P<\,-1$
$\therefore P \in\left(-\infty, -1\right)$