Question

Let the set of all values of $ p \in \mathbb{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:

• A. 0
• B. 9
• C. 5
• D. 20

Hint:

For quadratic equations, analyze the discriminant and the sum and product of the roots to determine the conditions under which the roots are real and satisfy other constraints.

Answer 1

The Correct Option is C

Given the inequalities: \[ p + 2 < 0 \Rightarrow p < -2 \] and \[ 2p + 9 > 0 \Rightarrow p > -\frac{9}{2} \]. For the discriminant \( D \ge 0 \), we have: \[ (p + 2)^2 - 4(2p + 9) \ge 0 \] \[ p^2 + 4p + 4 - 8p - 36 \ge 0 \] \[ p^2 - 4p - 32 \ge 0 \] Factoring the quadratic yields: \[ (p - 8)(p + 4) \ge 0 \] which implies \( p \in (-\infty, -4] \cup [8, \infty) \). Combining all conditions, the valid range for \( p \) is: \[ p \in \left[-\frac{9}{2}, -4\right] \] From this interval, we identify \( \alpha = -\frac{9}{2} \) and \( \beta = -4 \). Calculating \( \beta - 2\alpha \): \[ \beta - 2\alpha = -4 + 9 = 5 \] Thus, \( \boxed{\beta - 2\alpha = 5} \).