Let the set of all values of $p \in \mathbb{R}$, 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

<p>To find the set of all values of $ p \in \mathbb{R} $ for which both roots of the equation $ x^2-(p+2)x+(2p+9)=0 $ are negative real numbers, follow these steps:</p> <p><p><strong>Discriminant Condition</strong>: </p> <p>The equation's discriminant $ D $ must be non-negative for real roots:</p> <p>$ (p+2)^2 - 4(2p+9) \geq 0 $</p> <p>Simplifying this:</p> <p>$ p^2 + 4p + 4 - 8p - 36 \geq 0 \quad \Rightarrow \quad p^2 - 4p - 32 \geq 0 $</p> <p>This can be factored as:</p> <p>$ (p-8)(p+4) \geq 0 $</p> <p>Meaning $ p \in (-\infty, -4] \cup [8, \infty) $. …… (1)</p></p> <p><p><strong>Sum of Roots Condition</strong>: </p> <p>The sum of the roots (which is $ p+2 $) must be negative:</p> <p>$ p + 2 < 0 \quad \Rightarrow \quad p < -2 $ …… (2)</p></p> <p><p><strong>Product of Roots Condition</strong>:</p> <p>The product of the roots $ (2p + 9) $ must be positive:</p> <p>$ 2p + 9 > 0 \quad \Rightarrow \quad p > -\frac{9}{2} $ …… (3)</p></p> <p><p><strong>Determine the Valid Interval</strong>:</p> <p>Combine the results from conditions (1), (2), and (3). From conditions (1) and (2), we find $ p < -2 $:</p> <p>Intersection of $ (-\infty, -4] $ and $ (-\frac{9}{2}, -2) $ gives:</p> <p>$ p \in \left(-\frac{9}{2}, -4\right] $</p></p> <p><p><strong>Calculate $\beta - 2\alpha$</strong>:</p> <p>With $\alpha = -\frac{9}{2}$ and $\beta = -4$, compute:</p> <p>$ \beta - 2\alpha = -4 - 2\left(-\frac{9}{2}\right) = -4 + 9 = 5 $</p></p> <p>Therefore, the difference $\beta - 2\alpha$ is 5.</p>