The sum of all the roots of the equation (x-1)^2-5|x-1|+6=0 , is:

Solution & Explanation

### Related Formula X^2 = |X|^2 Quadratic factorization: t^2 - 5t + 6 = (t-2)(t-3) ### Core Logic Rewrite the equation taking |x - 1| = t, where t geq 0. Since (x-1)^2 = |x-1|^2, the equation becomes: t^2 - 5t + 6 = 0 ### Step 1: Solve for modulus (t - 2)(t - 3) = 0 Rightarrow t = 2, 3 Since both roots are positive, both provide valid solutions for the modulus. |x - 1| = 2 quad textand quad |x - 1| = 3 ### Step 2: Unpack x values From |x - 1| = 2: x - 1 = 2 Rightarrow x = 3 x - 1 = -2 Rightarrow x = -1 From |x - 1| = 3: x - 1 = 3 Rightarrow x = 4 x - 1 = -3 Rightarrow x = -2 The roots are 3, -1, 4, -2. ### Step 3: Sum of Roots textSum = 3 + (-1) + 4 + (-2) = 4 ### Pattern Recognition In symmetric modulus equations f(|x-a|) = 0, every valid root t generates twin solutions (a+t) and (a-t). The sum of each pair is perfectly 2a. If there are n distinct valid positive roots for t, the sum of all x-roots is exactly n times 2a. Here, n=2, a=1 Rightarrow 2 times 2(1) = 4. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Quadratic Equations

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Q57 jee_main_2025_03_april_evening Nature of Roots
Let the equation x(x + 2)(12 - k) = 2 have equal roots. Then the distance of the point (k, frack2) from the line 3x + 4y + 5 = 0 is
  • A. 15
  • B. 5sqrt3
  • C. 15sqrt5
  • D. 12

Solution

### Related Formula For a quadratic equation ax^2 + bx + c = 0 to have equal roots, its discriminant must be zero: D = b^2 - 4ac = 0 Perpendicular distance of point (x_0, y_0) from line Ax + By + C = 0 is: d = frac|Ax_0 + By_0 + C|sqrtA^2 + B^2 ### Core Logic Let's expand the given equation: (x^2 + 2x)(12 - k) = 2 Let lambda = 12-k. The quadratic equation is: lambda x^2 + 2lambda x - 2 = 0 quad (lambda neq 0) ### Step 1: Finding k Set the discriminant to zero: D = (2lambda)^2 - 4(lambda)(-2) = 0 4lambda^2 + 8lambda = 0 implies 4lambda(lambda + 2) = 0 Since lambda neq 0 (otherwise it is not quadratic and has no roots): lambda = -2 Thus: 12 - k = -2 implies k = 14 ### Step 2: Calculating perpendicular distance The point of interest is: (k, frack2) = (14, 7) Distance from the line 3x + 4y + 5 = 0: d = frac|3(14) + 4(7) + 5|sqrt3^2 + 4^2 = frac|42 + 28 + 5|5 = frac755 = 15 ### Pattern Recognition In quadratic equation analysis, substitution of variable coefficients with parameter lambda keeps calculations clean and helps identify constraints such as lambda neq 0 at early stages. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Quadratic Equations Class 10 Mathematics: Coordinate Geometry
Q64 jee_main_2025_07_april_morning Location of Roots
Let the set of all values of p in mathbbR , 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 - 2alpha is equal to
  • A. 0
  • B. 9
  • C. 5
  • D. 20

Solution

### Related Formula For both roots of a quadratic equation ax^2 + bx + c = 0 to be negative real numbers, three mandatory rules must be met simultaneously: 1. D ge 0 (Real roots) 2. Sum of roots = -b/a < 0 3. Product of roots = c/a > 0 ### Core Logic From the given quadratic equation x^2 - (p + 2)x + (2p + 9) = 0: **Condition 1**: Discriminant D 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 - 8)(p + 4) ge 0 implies p in (-infty, -4] cup [8, infty) quad dots (i) ### Step 1: Evaluate Sum and Product Conditions
Location of Roots diagram for Q64 - JEE Main 2025 Morning
Location of Roots diagram for Q64 - JEE Main 2025 Morning
**Condition 2**: Sum of roots < 0 alpha + beta = p + 2 < 0 implies p < -2 quad dots (ii) **Condition 3**: Product of roots > 0 alphabeta = 2p + 9 > 0 implies p > -frac92 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) - Intersecting this with (i) limits the range cleanly to: p in left(-frac92, -4right] Thus, alpha = -frac92 and beta = -4. ### Step 3: Final Value Calculation Calculate the requested target expression: beta - 2alpha = -4 - 2left(-frac92right) = -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
Q56 jee_main_2025_08_april_evening Equations Involving Absolute Value
The sum of the squares of the roots of |x + 2|^2 + |x - 2| - 2 = 0 and the squares of the roots of x^2 - 2|x - 3| - 5 = 0, is
  • A. 26
  • B. 36
  • C. 30
  • D. 24

Solution

### Related Formula textSum of squares of roots = (alpha + beta)^2 - 2alphabeta ### 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 Since |x-2| ge 0, we choose: |x-2| = 1 implies x = 3 text or 1 Sum of squares of roots = 3^2 + 1^2 = 10 ### Step 2: Evaluate Second Modulus Equation (Case Analysis) For x^2 - 2|x - 3| - 5 = 0: * Case I (x ge 3): x^2 - 2x + 6 - 5 = 0 implies (x-1)^2 = 0 implies x = 1 (Rejected since x ge 3). * Case II (x < 3): 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, roots satisfy validation checks. textSum of squares = (-2)^2 - 2(-11) = 4 + 22 = 26 textTotal 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
Q51 jee_main_2025_29_jan_evening Nature of Roots
If the set of all a in mathbbR, for which the equation 2x^2 + (a - 5)x + 15 = 3a has no real root, is the interval (alpha, beta), and X = \x in mathbbZ : alpha < x < beta\, then sum_x in X x^2 is equal to
  • A. 2109
  • B. 2129
  • C. 2139
  • D. 2119

Solution

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

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