Let the set of all values of p ∈ R, for which both the roots of the equation x² - (p+2)x + (2p+9) =

Q: Why do we use $D \ge 0$ instead of $D > 0$?

The problem specifies 'real numbers', not 'distinct real numbers'. If $D=0$, the roots are equal, which are still real and negative. Excluding the $D=0$ case would be a fatal error that loses you points.

Q: Why is the product of roots $P > 0$ necessary?

If the product $P < 0$, one root is positive and one is negative. Since we require *both* to be negative, their product must be positive (negative times negative equals positive).

Q: Can I just solve for $x$ using the quadratic formula?

While theoretically possible, it leads to a messy expression involving square roots of $p$. Using the properties of roots is the elegant, 'JEE-style' path that avoids algebraic nightmares and keeps your logic clean.

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

Options:

(A) 0

(B) 9

(C) 5

(D) 20

Sigma Percentile · Accepted Answer

### Analyzing the Setup For the quadratic equation $x^2 - (p+2)x + (2p+9) = 0$ to have two negative real roots, we must satisfy the "Holy Trinity" of conditions. For any quadratic $ax^2 + bx + c = 0$, these conditions are: 1. The discriminant $D = b^2 - 4ac \ge 0$ (for real roots). 2. The sum of roots $S = -\frac{b}{a} < 0$ (for negative sum). 3. The product of roots $P = \frac{c}{a} > 0$ (for same-sign roots). ### The Discriminant Battle We first evaluate the discriminant $D \ge 0$ with $a=1$, $b=-(p+2)$, and $c=2p+9$: $$D = (-(p+2))^2 - 4(1)(2p+9) \ge 0$$ Expanding the expression, we obtain: $$p^2 + 4p + 4 - 8p - 36 \ge 0$$ $$p^2 - 4p - 32 \ge 0$$ Factoring the quadratic yields $(p-8)(p+4) \ge 0$. Using the wavy curve method, we identify the valid region for $p$ as: $$p \in (-\infty, -4] \cup [8, \infty)$$ ### The Sum and Product Constraints Next, we analyze the sum of the roots $S = -\frac{b}{a} < 0$: $$-(-(p+2)) < 0 \Rightarrow p+2 < 0 \Rightarrow p < -2$$ Finally, we ensure the product of the roots $P = \frac{c}{a} > 0$: $$\frac{2p+9}{1} > 0 \Rightarrow 2p > -9 \Rightarrow p > -\frac{9}{2}$$ ### The Intersection of Truth We must now find the intersection of the three derived constraints: 1. $p \in (-\infty, -4] \cup [8, \infty)$ 2. $p < -2$ 3. $p > -\frac{9}{2}$ The intersection of $p > -\frac{9}{2}$ and $p < -2$ gives the interval $(-\frac{9}{2}, -2)$. Intersecting this with the discriminant condition $p \in (-\infty, -4] \cup [8, \infty)$, we find the valid range for $p$ is: $$p \in \left(-\frac{9}{2}, -4\right]$$ Comparing this to the given interval $(\alpha, \beta]$, we identify $\alpha = -\frac{9}{2}$ and $\beta = -4$. ### The Final Victory We are tasked with calculating the value of $\beta - 2\alpha$: $$\beta - 2\alpha = -4 - 2\left(-\frac{9}{2}\right)$$ $$-4 + 9 = 5$$ The final answer is **5**.

Animated Solution for Mathematics - Quadratic Equations: Let the set of all values of $p \in R$ , for which both the roots of the equation $x^{2} - (p + 2) x + (2 p + 9) = 0$ are negative real numbers, be the interval $(α, β]$ . Then $β - 2 α$ is equal to

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Visualized Solution

Problem Analysis

Conditions for Negative Real Roots

Condition 1: Discriminant $D \geq 0$

Solving the Discriminant Inequality

Interval for $D \geq 0$

Condition 2: Sum of Roots ( $S < 0$ )

Condition 3: Product of Roots ( $P > 0$ )

Finding the Common Intersection

Identifying $α$ and $β$

Final Calculation: $β - 2 α$

The Sigma Insight: Location of Roots

Analyzing the Setup

The Discriminant Battle

The Sum and Product Constraints

The Intersection of Truth

The Final Victory

Similar Questions

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Select Answer:

Visualized Solution

Problem Analysis

Conditions for Negative Real Roots

Condition 1: Discriminant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D≥0

Solving the Discriminant Inequality

Interval for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D≥0

Condition 2: Sum of Roots ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S<0)

Condition 3: Product of Roots ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P>0)

Finding the Common Intersection

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β−2α