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:
Show Hint
- 0
- 9
- 5
- 20
The Correct Option is C
Solution and Explanation
Given inequalities: \[ p + 2 < 0 \Rightarrow p < -2 \] \[ 2p + 9 > 0 \Rightarrow p > -\frac{9}{2} \] For the discriminant \( D \ge 0 \): \[ (p + 2)^2 - 4(2p + 9) \ge 0 \] \[ p^2 + 4p + 4 - 8p - 36 \ge 0 \] \[ p^2 - 4p - 32 \ge 0 \] \[ (p - 8)(p + 4) \ge 0 \] \[ p \in (-\infty, -4] \cup [8, \infty) \] Considering both conditions together: \[ p \in \left[-\frac{9}{2}, -4\right] \] Now, \[ \alpha = -\frac{9}{2}, \quad \beta = -4 \] \[ \beta - 2\alpha = -4 + 9 = 5 \] \[ \boxed{\beta - 2\alpha = 5} \]
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