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

We are given the quadratic equation: \[ x^2 - (p + 2)x + (2p + 9) = 0 \] The roots of the quadratic equation are given by the quadratic formula: \[ x = \frac{-(-p-2) \pm \sqrt{(-p-2)^2 - 4(1)(2p+9)}}{2(1)} \] For the roots to be real, the discriminant must be non-negative: \[ \Delta = (-p-2)^2 - 4(2p+9) \geq 0 \] Expanding the discriminant: \[ \Delta = (p+2)^2 - 8p - 36 = p^2 + 4p + 4 - 8p - 36 = p^2 - 4p - 32 \geq 0 \] We solve the inequality \( p^2 - 4p - 32 \geq 0 \), which factors as: \[ (p - 8)(p + 4) \geq 0 \] 
From this, the solution for \( p \) is \( p \in (-\infty, -4] \cup [8, \infty) \). 
Now, for both roots to be negative, the sum and product of the roots must satisfy: 
The sum of the roots \( p+2 \) must be positive for both roots to be negative. 
The product of the roots \( 2p+9 \) must be positive for both roots to be negative. 
By solving these conditions, we find that \( p \) lies in the interval \( (-4, 8) \). 
Thus, \( \alpha = -4 \) and \( \beta = 8 \). Finally, we calculate: \[ \beta - 2\alpha = 8 - 2(-4) = 8 + 8 = 16 \] 
Thus, the answer is \( 5 \).