Let the set of all values of p in R$p \in \mathbb{R}$ , for which both the roots of the equation x² - (p + 2)x + (2p + 9) = 0$x^2 - (p + 2)x + (2p + 9) = 0$ are negative real numbers, be the interval (α, β]$(\alpha, \beta]$ . Then β - 2α$\beta - 2\alpha$ is equal to

Related Formula

Core Logic

Examine both absolute value equations under interval tracking guidelines. Follow structural tracking limits from the source sheets to cleanly process algebraic paths without context deviations.

Step 1: Evaluate First Modulus Equation

Following reference solution steps for the localized structural format path:

Since |x-2| ≥ 0$|x-2| \ge 0$, we choose:

Sum of squares of roots = 3² + 1² = 10$= 3^2 + 1^2 = 10$

Step 2: Evaluate Second Modulus Equation (Case Analysis)

For x² - 2|x - 3| - 5 = 0$x^2 - 2|x - 3| - 5 = 0$:

• Case I (x ≥ 3$x \ge 3$): x² - 2x + 6 - 5 = 0 (x-1)² = 0 x = 1$x^2 - 2x + 6 - 5 = 0 \implies (x-1)^2 = 0 \implies x = 1$ (Rejected since x ≥ 3$x \ge 3$).
• Case II (x < 3$x < 3$): x² + 2x - 6 - 5 = 0 x² + 2x - 11 = 0$x^2 + 2x - 6 - 5 = 0 \implies x^2 + 2x - 11 = 0$

Step 3: Final Combined Calculation

For the acceptable equation x² + 2x - 11 = 0$x^2 + 2x - 11 = 0$, roots satisfy validation checks.

Pattern Recognition

Always double check constraints when switching intervals in modulus cases. A valid algebraic root is useless if it falls outside its defining condition boundary map.

Chapter Mix

Class 11 Mathematics: Quadratic Equations

Related Formula

For a quadratic equation Ax² + Bx + C = 0$Ax^2 + Bx + C = 0$ to have no real roots, its discriminant must be strictly negative:

Core Logic

Rearranging the given equation into standard quadratic form:

Here, A = 2$A = 2$, B = a - 5$B = a - 5$, and C = 15 - 3a$C = 15 - 3a$. Setting the discriminant less than zero:

Step 1: Solve for the Interval

Factorizing the quadratic inequality:

Thus, a in (-19, 5)$a \in (-19, 5)$. This gives α = -19$\alpha = -19$ and \beta = 5.

Step 2: Calculate the Sum of Squares

The set X$X$ consists of integers strictly between -19$-19$ and 5$5$:

Pattern Recognition

Recognize that the negative terms squared are identical to the positive terms squared. Splitting the summation avoids calculating large numbers manually or allows using standard formula templates like (n(n+1)(2n+1))/(6)$\frac{n(n+1)(2n+1)}{6}$ efficiently.

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Class 11 Mathematics: Quadratic Equations Class 11 Mathematics: Sequences and Series

Related Formula

Modulus definition rule:

Core Logic

Analyze the roots by splitting into cases around the critical threshold x = (3)/(2)$x = \frac{3}{2}$:

Case I: x ≥ (3)/(2)$x \ge \frac{3}{2}$

(We select the positive root since 2√(2)-1 ≥ 1.5$2\sqrt{2}-1 \ge 1.5$).

Step 1: Evaluating the alternate domain branch

Case II: x < (3)/(2)$x < \frac{3}{2}$

(We select 1-√(2)$1-\sqrt{2}$ since it satisfies the inequality constraint).

Step 2: Summing the Squares of the Roots

Pattern Recognition

Always validate absolute root values against their domain restrictions to avoid including phantom solutions.

Chapter Mix

Class 11 Maths: Quadratic Equations

Related Formula

Newton's Theorem for roots: If α, β$\alpha, \beta$ satisfy ax² + bx + c = 0$ax^2 + bx + c = 0$ and Sₙ = αⁿ + βⁿ$S_n = \alpha^n + \beta^n$, then:

Core Logic

Apply Newton's Theorem directly to the first equation x² + √(3)x - 16 = 0$x^2 + \sqrt{3}x - 16 = 0$:

For n = 25$n = 25$:

Dividing both sides by 2P₂₃$2P_{23}$:

Step 1: Evaluation of the Second Part

For the second equation x² + 3x - 1 = 0$x^2 + 3x - 1 = 0$:

For n = 25$n = 25$:

Dividing both sides by Q₂₄$Q_{24}$:

Step 2: Total Calculation

Add both evaluated components:

Pattern Recognition

Shortcut: High sequential indices (25, 24, 23$25, 24, 23$) indicate recurrence via Newton's Theorem. Relate Pₙ$P_n$ and Qₙ$Q_n$ directly to their characteristic quadratic polynomials to evaluate the ratios in one step without calculating powers.

Evaluation Rubric / Model Answer

Option (C)

Chapter Mix

Class 11 Mathematics: Quadratic Equations