The value of $p$ for which both roots of the quadratic equation $4 x^{2} - 20 P x + (25 P^{2} + 15 P - 66) = 0$ are less than $2$ , lies in

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Solution

Given,

$4 x^{2} - 20 P x + (25 P^{2} + 15 P - 66) = 0$

$D \geq 0$

$\Rightarrow (- 20 p)^{2} - 4 \times 4 (25 p^{2} + 15 p - 66) \geq 0$

$16 [25 p^{2} - 25 p^{2} - 15 p + 66] \geq 0$

$- 15 p + 66 \geq 0$

$15 p - 66 \leq 0$

$p \leq \frac{66}{15}$

$\Rightarrow - \frac{b}{2 a} < 2$

$\frac{20 p}{8} < 2$

$p < \frac{16}{20}$

$\Rightarrow p < \frac{4}{5}$

$f (2) > 0$

$16 - 40 p + 25 p^{2} + 15 p - 66 > 0$

$\Rightarrow 25 p^{2} - 25 p - 50 > 0$

$p^{2} - p - 2 > 0$

$(p - 2) (p + 1) > 0$

$p \in (- \infty, - 1) \cup (2, \infty)$

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