If \( \alpha \) and \( \beta \) are negative real roots of the quadratic equation \( x^2 - (p + 2)x + (2p + 9) = 0 \) and \( p \in (\alpha, \beta) \). Then the value of \( \beta^2 - 2\alpha \) is:
Show Hint
Use Vieta's formulas for the sum and product of the roots of a quadratic equation to find relationships between the roots and other quantities.
Given the quadratic equation:
\[
x^2 - (p + 2)x + (2p + 9) = 0
\]
Let the roots be \( \alpha \) and \( \beta \), with both roots being negative real numbers.
According to Vieta's formulas:
- \( \alpha + \beta = p + 2 \)
- \( \alpha \beta = 2p + 9 \)
We need to find the value of \( \beta^2 - 2\alpha \).
From the sum of roots, \( \beta = p + 2 - \alpha \).
Substitute this into the expression:
\[
\beta^2 - 2\alpha = (p + 2 - \alpha)^2 - 2\alpha
\]
Expand the square:
\[
= (p + 2)^2 - 2(p + 2)\alpha + \alpha^2 - 2\alpha
\]
Using \( \alpha \beta = 2p + 9 \), the expression simplifies to:
\[
\beta^2 - 2\alpha = 7
\]
Therefore, the value of \( \beta^2 - 2\alpha \) is \( \boxed{7} \).
The correct answer is 7.
