The product of all the rational roots of the equation (x² - 9x + 11)² - (x - 4)(x - 5) = 3$(x^{2} - 9x + 11)^{2} - (x - 4)(x - 5) = 3$ is equal to :

Core Logic

For a polynomial P(x) = x³ + ax² + bx + c$P(x) = x^3 + ax^2 + bx + c$ to be divisible by x² + 2$x^2 + 2$, we can perform polynomial long division or use synthetic substitution (roots of x²+2=0$x^2+2=0$). Alternatively, factorize P(x)$P(x)$: Since the degree is 3 and the leading coefficient is 1, we must have:

where k$k$ is a constant.

Step 1: Equating Coefficients

Expand the right side:

Comparing coefficients with x³ + ax² + bx + c$x^3 + ax^2 + bx + c$: x²$x^2$ coefficient: a = k$a = k$ x$x$ coefficient: b = 2$b = 2$ Constant term: c = 2k$c = 2k$

Step 2: Apply Constraints

From the coefficient matching, we have a = (c)/(2)$a = \frac{c}{2}$ and b = 2$b = 2$. Since a, b, c in N$a, b, c \in \mathbb{N}$ and a, b, c ≤ 20$a, b, c \leq 20$: b = 2$b = 2$ (this is fixed, always valid). c$c$ must be an even natural number such that c ≤ 20$c \leq 20$. The possible values for c$c$ are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20$\{2, 4, 6, 8, 10, 12, 14, 16, 18, 20\}$. For each such c$c$, a$a$ is uniquely determined as a = c/2$a = c/2$ and a ≤ 10$a \leq 10$ (which easily satisfies a ≤ 20$a \leq 20$).

Step 3: Final Count

The number of valid (a, b, c)$(a, b, c)$ tuples corresponds to the number of valid c$c$ values. There are exactly 10$10$ values for c$c$. Thus, the number of polynomials in S$S$ is 10$10$.

Chapter Mix

Class 10 Mathematics: Polynomials Class 11 Mathematics: Complex Numbers and Quadratic Equations

Related Formula

For a quadratic equation ax² + bx + c = 0$ax^2 + bx + c = 0$ to have equal roots, its discriminant must be zero:

Perpendicular distance of point (x₀, y₀)$(x_0, y_0)$ from line Ax + By + C = 0$Ax + By + C = 0$ is:

Core Logic

Let's expand the given equation:

Let λ = 12-k$\lambda = 12-k$. The quadratic equation is:

Step 1: Finding k$k$

Set the discriminant to zero:

Since λ ≠ 0$\lambda \neq 0$ (otherwise it is not quadratic and has no roots): λ = -2$\lambda = -2$

Thus:

Step 2: Calculating perpendicular distance

The point of interest is:

Distance from the line 3x + 4y + 5 = 0$3x + 4y + 5 = 0$:

Pattern Recognition

In quadratic equation analysis, substitution of variable coefficients with parameter λ$\lambda$ keeps calculations clean and helps identify constraints such as λ ≠ 0$\lambda \neq 0$ at early stages.

Chapter Mix

Class 11 Mathematics: Quadratic Equations Class 10 Mathematics: Coordinate Geometry

Related Formula

For both roots of a quadratic equation ax² + bx + c = 0$ax^2 + bx + c = 0$ to be negative real numbers, three mandatory rules must be met simultaneously:

• D ≥ 0$D \ge 0$ (Real roots)
• Sum of roots = -b/a < 0$= -b/a < 0$
• Product of roots = c/a > 0$= c/a > 0$

Core Logic

From the given quadratic equation x² - (p + 2)x + (2p + 9) = 0$x^2 - (p + 2)x + (2p + 9) = 0$:

Condition 1: Discriminant D ≥ 0$D \ge 0$

Step 1: Evaluate Sum and Product Conditions

Condition 3: Product of roots > 0$> 0$

Step 2: Find Intersection Domain

Take the operational intersection across all three parameters: (i), (ii), and (iii):

• From (ii) and (iii): p in (-(9)/(2), -2)$p \in \left(-\frac{9}{2}, -2\right)$
• Intersecting this with (i) limits the range cleanly to:

Thus, α = -(9)/(2)$\alpha = -\frac{9}{2}$ and β = -4$\beta = -4$.

Step 3: Final Value Calculation

Calculate the requested target expression:

Pattern Recognition

Remember that if roots are strictly real and matching signs, managing product rules before analyzing spatial configurations saves major compute overhead during intersection evaluation.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations

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.

Chapter Mix

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