Let the set of all values of $\mathrm{p} \in \mathbb{R}$, for which both the roots of the equation $x^2-(p+2) x+(2 p+9)=0$ are negative real numbers, be the interval $(\alpha, \beta]$. Then $\beta-2 \alpha$ is equal to

• A

  0
• B

  9
• Correct answer
  C

  5
• D

  20

Solution

Using location of roots :

(i) $\mathrm{D} \geq 0$
(ii) $\frac{-\mathrm{b}}{2 \mathrm{a}} \lt 0$
(iii) a. $\mathrm{f}(0) \gt 0$
$\begin{aligned}<br/>& (p+2)^2-4(2 p+9) \geq 0 \\ & (p+4)(p-8) \geq 0 \quad p+2 \lt 0 \quad 2 p+9 \gt 0<br/>\end{aligned}$
Intersection $p \in\left(-\frac{9}{2},-4\right]$
$\therefore \beta-2 \alpha=-4+9=5$

Hence, the correct answer is option: 'C'.