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(C) For the roots of the quadratic equation $x^2-(p+2)x+(2p+9)=0$ to be negative real numbers,the following conditions must be satisfied:$1$. The disc

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$5$

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(C) For the roots of the quadratic equation $x^2-(p+2)x+(2p+9)=0$ to be negative real numbers,the following conditions must be satisfied:$1$. The discriminant $D \geq 0$:$D = (p+2)^2 - 4(2p+9) \geq 0$$p^2 + 4p + 4 - 8p - 36 \geq 0$$p^2 - 4p - 32 \geq 0$$(p-8)(p+4) \geq 0$This implies $p \in (-\infty, -4] \cup [8, \infty)$.$2$. The sum of roots must be negative:Sum $= -\frac{b}{a} = p+2 $3$. The product of roots must be positive:Product $= \frac{c}{a} = 2p+9 > 0 \implies p > -\frac{9}{2}$.Combining these conditions: $p \in (-\frac{9}{2}, -4]$.Thus,$\alpha = -\frac{9}{2}$ and $\beta = -4$.We need to find $\beta - 2\alpha = -4 - 2(-\frac{9}{2}) = -4 + 9 = 5$.

Column-$I$	Column-$II$
$A$. Imaginary roots	$P$. $a \in (-1, 4)$
$B$. One root less than $3$ and other greater than $3$	$Q$. $a \in (-\infty, -1)$
$C$. One root less than $1$ and other greater than $3$	$R$. $a \in (-\infty, -4/3)$

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

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