### Related Formula For a quadratic equation ax^2 + bx + c = 0$ax^2 + bx + c = 0$ to have equal roots, its discriminant must be zero: D = b^2 - 4ac = 0$D = b^2 - 4ac = 0$ Perpendicular distance of point (x_0, y_0)$(x_0, y_0)$ from line Ax + By + C = 0$Ax + By + C = 0$ is: d = frac|Ax_0 + By_0 + C|sqrtA^2 + B^2$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$ ### Core Logic Let's expand the given equation: (x^2 + 2x)(12 - k) = 2$(x^2 + 2x)(12 - k) = 2$ Let lambda = 12-k$\lambda = 12-k$. The quadratic equation is: lambda x^2 + 2lambda x - 2 = 0 quad (lambda neq 0)$\lambda x^2 + 2\lambda x - 2 = 0 \quad (\lambda \neq 0)$ ### Step 1: Finding k$k$ Set the discriminant to zero: D = (2lambda)^2 - 4(lambda)(-2) = 0$D = (2\lambda)^2 - 4(\lambda)(-2) = 0$ 4lambda^2 + 8lambda = 0 implies 4lambda(lambda + 2) = 0$4\lambda^2 + 8\lambda = 0 \implies 4\lambda(\lambda + 2) = 0$ Since lambda neq 0$\lambda \neq 0$ (otherwise it is not quadratic and has no roots): lambda = -2$\lambda = -2$ Thus: 12 - k = -2 implies k = 14$12 - k = -2 \implies k = 14$ ### Step 2: Calculating perpendicular distance The point of interest is: (k, frack2) = (14, 7)$(k, \frac{k}{2}) = (14, 7)$ Distance from the line 3x + 4y + 5 = 0$3x + 4y + 5 = 0$: d = frac|3(14) + 4(7) + 5|sqrt3^2 + 4^2 = frac|42 + 28 + 5|5 = frac755 = 15$d = \frac{|3(14) + 4(7) + 5|}{\sqrt{3^2 + 4^2}} = \frac{|42 + 28 + 5|}{5} = \frac{75}{5} = 15$ ### Pattern Recognition In quadratic equation analysis, substitution of variable coefficients with parameter lambda$\lambda$ keeps calculations clean and helps identify constraints such as lambda neq 0$\lambda \neq 0$ at early stages. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Quadratic Equations Class 10 Mathematics: Coordinate Geometry

### Related Formula For both roots of a quadratic equation ax^2 + bx + c = 0$ax^2 + bx + c = 0$ to be negative real numbers, three mandatory rules must be met simultaneously: 1. D ge 0$D \ge 0$ (Real roots) 2. Sum of roots = -b/a < 0$= -b/a < 0$ 3. Product of roots = c/a > 0$= c/a > 0$ ### Core Logic From the given quadratic equation x^2 - (p + 2)x + (2p + 9) = 0$x^2 - (p + 2)x + (2p + 9) = 0$: **Condition 1**: Discriminant D ge 0$D \ge 0$ D = [-(p + 2)]^2 - 4(1)(2p + 9) ge 0$D = [-(p + 2)]^2 - 4(1)(2p + 9) \ge 0$ p^2 + 4p + 4 - 8p - 36 ge 0 implies p^2 - 4p - 32 ge 0$p^2 + 4p + 4 - 8p - 36 \ge 0 \implies p^2 - 4p - 32 \ge 0$ (p - 8)(p + 4) ge 0 implies p in (-infty, -4] cup [8, infty) quad dots (i)$(p - 8)(p + 4) \ge 0 \implies p \in (-\infty, -4] \cup [8, \infty) \quad \dots (i)$ ### Step 1: Evaluate Sum and Product Conditions

**Condition 2**: Sum of roots < 0$< 0$ alpha + beta = p + 2 < 0 implies p < -2 quad dots (ii)$\alpha + \beta = p + 2 < 0 \implies p < -2 \quad \dots (ii)$ **Condition 3**: Product of roots > 0$> 0$ alphabeta = 2p + 9 > 0 implies p > -frac92 quad dots (iii)$\alpha\beta = 2p + 9 > 0 \implies p > -\frac{9}{2} \quad \dots (iii)$ ### Step 2: Find Intersection Domain Take the operational intersection across all three parameters: (i), (ii), and (iii): - From (ii) and (iii): p in left(-frac92, -2right)$p \in \left(-\frac{9}{2}, -2\right)$ - Intersecting this with (i) limits the range cleanly to: p in left(-frac92, -4right]$p \in \left(-\frac{9}{2}, -4\right]$ Thus, alpha = -frac92$\alpha = -\frac{9}{2}$ and beta = -4$\beta = -4$. ### Step 3: Final Value Calculation Calculate the requested target expression: beta - 2alpha = -4 - 2left(-frac92right) = -4 + 9 = 5$\beta - 2\alpha = -4 - 2\left(-\frac{9}{2}\right) = -4 + 9 = 5$ ### 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers and Quadratic Equations

### Related Formula textSum of squares of roots = (alpha + beta)^2 - 2alphabeta$\text{Sum of squares of roots} = (\alpha + \beta)^2 - 2\alpha\beta$ ### 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: |x-2|^2 + 2|x-2| - |x-2| - 2 = 0 implies (|x-2|+2)(|x-2|-1) = 0$|x-2|^2 + 2|x-2| - |x-2| - 2 = 0 \implies (|x-2|+2)(|x-2|-1) = 0$ Since |x-2| ge 0$|x-2| \ge 0$, we choose: |x-2| = 1 implies x = 3 text or 1$|x-2| = 1 \implies x = 3 \text{ or } 1$ Sum of squares of roots = 3^2 + 1^2 = 10$= 3^2 + 1^2 = 10$ ### Step 2: Evaluate Second Modulus Equation (Case Analysis) For x^2 - 2|x - 3| - 5 = 0$x^2 - 2|x - 3| - 5 = 0$: * Case I (x ge 3$x \ge 3$): x^2 - 2x + 6 - 5 = 0 implies (x-1)^2 = 0 implies x = 1$x^2 - 2x + 6 - 5 = 0 \implies (x-1)^2 = 0 \implies x = 1$ (Rejected since x ge 3$x \ge 3$). * Case II (x < 3$x < 3$): x^2 + 2x - 6 - 5 = 0 implies x^2 + 2x - 11 = 0$x^2 + 2x - 6 - 5 = 0 \implies x^2 + 2x - 11 = 0$ ### Step 3: Final Combined Calculation For the acceptable equation x^2 + 2x - 11 = 0$x^2 + 2x - 11 = 0$, roots satisfy validation checks. textSum of squares = (-2)^2 - 2(-11) = 4 + 22 = 26$\text{Sum of squares} = (-2)^2 - 2(-11) = 4 + 22 = 26$ textTotal Combined Value = 10 + 26 = 36$\text{Total Combined Value} = 10 + 26 = 36$ ### 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Quadratic Equations

### Related Formula For a quadratic equation Ax^2 + Bx + C = 0$Ax^2 + Bx + C = 0$ to have no real roots, its discriminant must be strictly negative: D = B^2 - 4AC < 0$D = B^2 - 4AC < 0$ ### Core Logic Rearranging the given equation into standard quadratic form: 2x^2 + (a - 5)x + (15 - 3a) = 0$2x^2 + (a - 5)x + (15 - 3a) = 0$ Here, A = 2$A = 2$, B = a - 5$B = a - 5$, and C = 15 - 3a$C = 15 - 3a$. Setting the discriminant less than zero: (a - 5)^2 - 4(2)(15 - 3a) < 0$(a - 5)^2 - 4(2)(15 - 3a) < 0$ (a^2 - 10a + 25) - 8(15 - 3a) < 0$(a^2 - 10a + 25) - 8(15 - 3a) < 0$ a^2 - 10a + 25 - 120 + 24a < 0$a^2 - 10a + 25 - 120 + 24a < 0$ a^2 + 14a - 95 < 0$a^2 + 14a - 95 < 0$ ### Step 1: Solve for the Interval Factorizing the quadratic inequality: (a + 19)(a - 5) < 0$(a + 19)(a - 5) < 0$ Thus, a in (-19, 5)$a \in (-19, 5)$. This gives alpha = -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$5$: X = \-18, -17, dots, 0, 1, 2, 3, 4\$X = \{-18, -17, \dots, 0, 1, 2, 3, 4\}$ sum_x in X x^2 = (-18)^2 + (-17)^2 + dots + 4^2$\sum_{x \in X} x^2 = (-18)^2 + (-17)^2 + \dots + 4^2$ = (1^2 + 2^2 + 3^2 + 4^2) + (1^2 + 2^2 + dots + 18^2)$= (1^2 + 2^2 + 3^2 + 4^2) + (1^2 + 2^2 + \dots + 18^2)$ = frac4 times 5 times 96 + frac18 times 19 times 376$= \frac{4 \times 5 \times 9}{6} + \frac{18 \times 19 \times 37}{6}$ = 30 + 2109 = 2139$= 30 + 2109 = 2139$ ### 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 fracn(n+1)(2n+1)6$\frac{n(n+1)(2n+1)}{6}$ efficiently. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Quadratic Equations Class 11 Mathematics: Sequences and Series

### Related Formula Modulus definition rule: |x| = begincases x, & x ge 0 \\ -x, & x < 0 endcases$|x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases}$ ### Core Logic Analyze the roots by splitting into cases around the critical threshold x = frac32$x = \frac{3}{2}$: **Case I:** x ge frac32$x \ge \frac{3}{2}$ x^2 + 2x - 3 - 4 = 0 implies x^2 + 2x - 7 = 0 implies x = 2sqrt2 - 1$x^2 + 2x - 3 - 4 = 0 \implies x^2 + 2x - 7 = 0 \implies x = 2\sqrt{2} - 1$ (We select the positive root since 2sqrt2-1 ge 1.5$2\sqrt{2}-1 \ge 1.5$). ### Step 1: Evaluating the alternate domain branch **Case II:** x < frac32$x < \frac{3}{2}$ x^2 - (2x - 3) - 4 = 0 implies x^2 - 2x - 1 = 0 implies x = 1 - sqrt2$x^2 - (2x - 3) - 4 = 0 \implies x^2 - 2x - 1 = 0 \implies x = 1 - \sqrt{2}$ (We select 1-sqrt2$1-\sqrt{2}$ since it satisfies the inequality constraint). ### Step 2: Summing the Squares of the Roots textSum of Squares = (2sqrt2 - 1)^2 + (1 - sqrt2)^2$\text{Sum of Squares} = (2\sqrt{2} - 1)^2 + (1 - \sqrt{2})^2$ = (8 - 4sqrt2 + 1) + (1 - 2sqrt2 + 2) = 12 - 6sqrt2 = 6(2 - sqrt2)$= (8 - 4\sqrt{2} + 1) + (1 - 2\sqrt{2} + 2) = 12 - 6\sqrt{2} = 6(2 - \sqrt{2})$ ### Pattern Recognition Always validate absolute root values against their domain restrictions to avoid including phantom solutions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Quadratic Equations