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

### Core Logic For the denominator x^2 - 8x + 32$x^2 - 8x + 32$, D = 64 - 128 < 0$D = 64 - 128 < 0$ and a = 1 > 0$a = 1 > 0$. Thus, x^2 - 8x + 32 > 0 forall x in mathbbR$x^2 - 8x + 32 > 0 \forall x \in \mathbb{R}$. ### Step 1: Constraint on Numerator Since the denominator is always positive, the numerator must be strictly negative for all x in mathbbR$x \in \mathbb{R}$. ax^2 + 2(a + 1)x + 9a + 4 < 0 quad forall x in mathbbR$$ax^2 + 2(a + 1)x + 9a + 4 < 0 \quad \forall x \in \mathbb{R}$$ This requires a < 0$a < 0$ and D < 0$D < 0$. ### Step 2: Conclusion Since a$a$ must be strictly less than 0$0$, there are no *positive* integral values of a$a$ that satisfy the condition. Hence, S$S$ is an empty set. Number of elements is 0. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Quadratic Equations