Let the set of all values of p R, for which both the roots of the equation x^2-(p+2) x+(2 …

Question

Let the set of all values of $p \in \mathbb{R}$, for which both the roots of the equation $x^2-(p+2) x+(2 p+9)=0$ are negative real numbers, be the interval $(\alpha, \beta]$. Then $\beta-2 \alpha$ is equal to

Accepted Answer

To find the set of all values of $ p \in \mathbb{R} $ for which both roots of the equation $ x^2-(p+2)x+(2p+9)=0 $ are negative real numbers, follow these steps:

Discriminant Condition:

The equation's discriminant $ D $ must be non-negative for real roots:

$    (p+2)^2 - 4(2p+9) \geq 0     $

Simplifying this:

$    p^2 + 4p + 4 - 8p - 36 \geq 0 \quad \Rightarrow \quad p^2 - 4p - 32 \geq 0     $

This can be factored as:

$    (p-8)(p+4) \geq 0     $

Meaning $ p \in (-\infty, -4] \cup [8, \infty) $.  \ldots \ldots  (1)

Sum of Roots Condition:

The sum of the roots (which is $ p+2 $) must be negative:

$    p + 2

Product of Roots Condition:

The product of the roots $ (2p + 9) $ must be positive:

$    2p + 9 > 0 \quad \Rightarrow \quad p > -\frac{9}{2}     $ \ldots \ldots  (3)

Determine the Valid Interval:

Combine the results from conditions (1), (2), and (3). From conditions (1) and (2), we find $ p

Calculate $\beta - 2\alpha$:

With $\alpha = -\frac{9}{2}$ and $\beta = -4$, compute:

$    \beta - 2\alpha = -4 - 2\left(-\frac{9}{2}\right) = -4 + 9 = 5    $

Therefore, the difference $\beta - 2\alpha$ is 5.

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