Key Concepts and Formulas

• For a quadratic equation $ax^2 + bx + c = 0$ to have real roots, the discriminant $D = b^2 - 4ac \ge 0$.
• If the roots $\alpha$ and $\beta$ of $ax^2 + bx + c = 0$ are both negative, then $\alpha + \beta = -\frac{b}{a} < 0$ and $\alpha \beta = \frac{c}{a} > 0$.

Step-by-Step Solution

Step 1: Write down the given quadratic equation and identify coefficients

• What and Why: We are given the quadratic equation and we need to identify the coefficients $a$, $b$, and $c$ to apply the conditions for negative real roots.
• Calculation: The given quadratic equation is $x^2 - (p+2)x + (2p+9) = 0$. Comparing with the standard form $ax^2 + bx + c = 0$, we have: $a = 1$ $b = -(p+2)$ $c = 2p+9$

Step 2: Apply the Discriminant Condition ($D \ge 0$)

• What and Why: The discriminant condition ensures that the roots are real. If $D < 0$, the roots are complex.
• Calculation: $D = b^2 - 4ac = [-(p+2)]^2 - 4(1)(2p+9) \ge 0$ $(p+2)^2 - 4(2p+9) \ge 0$ $p^2 + 4p + 4 - 8p - 36 \ge 0$ $p^2 - 4p - 32 \ge 0$ $(p-8)(p+4) \ge 0$ The roots of the quadratic $p^2 - 4p - 32 = 0$ are $p = 8$ and $p = -4$. Since the coefficient of $p^2$ is positive, the parabola opens upwards. Therefore, the inequality $(p-8)(p+4) \ge 0$ is satisfied when $p \le -4$ or $p \ge 8$.
• Result (Condition 1): $p \in (-\infty, -4] \cup [8, \infty) \quad \ldots(1)$

Step 3: Apply the Sum of Roots Condition ($S < 0$)

• What and Why: The sum of the roots must be negative for both roots to be negative.
• Calculation: The sum of the roots is given by $S = -\frac{b}{a} = -\frac{-(p+2)}{1} = p+2$. We require $S < 0$, so $p+2 < 0$ $p < -2$
• Result (Condition 2): $p \in (-\infty, -2) \quad \ldots(2)$

Step 4: Apply the Product of Roots Condition ($P > 0$)

• What and Why: The product of the roots must be positive for both roots to have the same sign. Since the roots are negative, their product is positive.
• Calculation: The product of the roots is given by $P = \frac{c}{a} = \frac{2p+9}{1} = 2p+9$. We require $P > 0$, so $2p+9 > 0$ $2p > -9$ $p > -\frac{9}{2}$
• Result (Condition 3): $p \in \left(-\frac{9}{2}, \infty\right) \quad \ldots(3)$

Step 5: Find the Intersection of the Three Intervals

• What and Why: We need to find the values of $p$ that satisfy all three conditions simultaneously. This means finding the intersection of the intervals from (1), (2), and (3).

• Calculation: We have the following intervals: (1) $p \in (-\infty, -4] \cup [8, \infty)$ (2) $p \in (-\infty, -2)$ (3) $p \in \left(-\frac{9}{2}, \infty\right)$

  First, intersect (2) and (3): $(-\infty, -2) \cap \left(-\frac{9}{2}, \infty\right) = \left(-\frac{9}{2}, -2\right)$ Since $-\frac{9}{2} = -4.5$, this interval is $(-4.5, -2)$.

  Now, intersect this result with (1): $\left(-\frac{9}{2}, -2\right) \cap \left( (-\infty, -4] \cup [8, \infty) \right)$

  The interval $(-4.5, -2)$ overlaps with $(-\infty, -4]$ in the region $(-4.5, -4]$. There is no overlap with $[8,\infty)$.

  Therefore, the common interval for $p$ that satisfies all conditions is: $p \in \left(-\frac{9}{2}, -4\right]$ So, $\alpha = -\frac{9}{2}$ and $\beta = -4$.

Step 6: Calculate $\beta - 2\alpha$

• What and Why: We need to calculate the value of the expression $\beta - 2\alpha$ using the values of $\alpha$ and $\beta$ we found.
• Calculation: $\beta - 2\alpha = -4 - 2\left(-\frac{9}{2}\right) = -4 + 9 = 5$

Common Mistakes & Tips

• Remember to consider all three conditions: discriminant, sum of roots, and product of roots. Forgetting one will lead to an incorrect interval.
• When solving inequalities, pay attention to the direction of the inequality sign, especially when multiplying or dividing by a negative number.
• Drawing a number line can be helpful in visualizing the intervals and finding their intersection.

Summary

To find the set of values of $p$ for which the quadratic equation $x^2-(p+2)x+(2p+9)=0$ has both roots as negative real numbers, we applied the conditions for the discriminant ($D \ge 0$), sum of roots ($S < 0$), and product of roots ($P > 0$). We found the intersection of the resulting intervals to determine the valid interval for $p$ as $\left(-\frac{9}{2}, -4\right]$. Finally, we calculated $\beta - 2\alpha = 5$.

Final Answer

The final answer is \boxed{5}, which corresponds to option (A).