Let the set of all values of $p \in R$, for which both the roots of theequation $x^2 - (p + 2)x + (2p+9)= 0$ are negative real numbers, be the interval $(\alpha, \beta)$. Then$\beta - 2\alpha$ is equal to
Finding Parameter p for Negative Quadratic Roots
We are given the quadratic equation $x^2 - (p + 2)x + (2p+9)= 0$. For both roots to be negative real numbers, three conditions must be met:
- The discriminant ($\Delta$) must be non-negative ($\Delta \ge 0$).
- The sum of the roots must be negative.
- The product of the roots must be positive.
Applying Conditions to the Equation
For the equation $ax^2 + bx + c = 0$, we have $a=1$, $b=-(p+2)$, and $c=(2p+9)$.
1. Discriminant Condition ($\Delta \ge 0$)
Calculate the discriminant $\Delta = b^2 - 4ac$:
$\Delta = (-(p+2))^2 - 4(1)(2p+9)$
$\Delta = (p+2)^2 - 4(2p+9)$
$\Delta = (p^2 + 4p + 4) - (8p + 36)$
$\Delta = p^2 - 4p - 32$
We need $\Delta \ge 0$, so $p^2 - 4p - 32 \ge 0$. Factoring the quadratic gives:
$(p-8)(p+4) \ge 0$
This inequality holds when $p \le -4$ or $p \ge 8$. So, $p \in (-\infty, -4] \cup [8, \infty)$.
2. Sum of Roots Condition (< $-\mathbf{0}$>)
The sum of the roots is given by $-b/a$. In this case, the sum is $\frac{-(-(p+2))}{1} = p+2$. For negative roots, the sum must be negative:
$p+2 < 0 \implies p < -2$. So, $p \in (-\infty, -2)$.
3. Product of Roots Condition (> $\mathbf{0}$>)
The product of the roots is given by $c/a$. In this case, the product is $\frac{2p+9}{1} = 2p+9$. For negative roots, the product must be positive:
$2p+9 > 0 \implies 2p > -9 \implies p > -9/2$ or $p > -4.5$. So, $p \in (-4.5, \infty)$.
Combining Conditions and Finding the Interval
We need to find the values of $p$ that satisfy all three conditions simultaneously:
- $p \in (-\infty, -4] \cup [8, \infty)$
- $p \in (-\infty, -2)$
- $p \in (-4.5, \infty)$
Let's find the intersection:
- Intersecting $(-\infty, -4] \cup [8, \infty)$ with $(-\infty, -2)$ gives $(-\infty, -4]$.
- Intersecting $(-\infty, -4]$ with $(-4.5, \infty)$ gives $(-4.5, -4]$.
The set of all values for $p$ is the interval $(-4.5, -4]$.
Calculating $\beta - 2\alpha$
The problem states the interval is $(\alpha, \beta]$. Comparing this with our result $(-4.5, -4]$, we have:
- $\alpha = -4.5$
- $\beta = -4$
Now, we calculate $\beta - 2\alpha$:
$\beta - 2\alpha = -4 - 2(-4.5)$
= $-4 - (-9)$
= $-4 + 9$
= $5$
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