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

Question

Let the set of all values of $p\in R$, 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

Accepted Answer

To find the set of all values of $p\in R$ 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:

$(p+2{)}^{2}-4(2p+9)\ge 0$

Simplifying this:

${p}^{2}+4p+4-8p-36\ge 0\, \Rightarrow \, {p}^{2}-4p-32\ge 0$

This can be factored as:

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

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+2Product of Roots Condition:

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

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

Determine the Valid Interval:

Combine the results from conditions (1), (2), and (3). From conditions (1) and (2), we find $pIntersection of $(-\infty,-4]$ and $(-\frac{9}{2},-2)$ gives:

$p\in (-\frac{9}{2},-4]$

Calculate $\beta -2\alpha $:

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

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

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

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