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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Q59 jee_main_2025_29_jan_morning Equations Reducible to Quadratic Forms
The number of solutions of the equation left(frac9x -frac9sqrtx +2right)left(frac2x -frac7sqrtx +3right) = 0 is:
  • A. 2
  • B. 4
  • C. 1
  • D. 3

Solution

### Related Formula textSubstitute variable to convert non-linear form: alpha = frac1sqrtx quad (x > 0) ### Core Logic Let frac1sqrtx = alpha. The equation reduces to a product of two quadratics: (9alpha^2 - 9alpha + 2)(2alpha^2 - 7alpha + 3) = 0 ### Step 1: Factorize the components First quadratic: 9alpha^2 - 9alpha + 2 = 0 implies (3alpha - 2)(3alpha - 1) = 0 implies alpha = frac23, frac13 Second quadratic: 2alpha^2 - 7alpha + 3 = 0 implies (2alpha - 1)(alpha - 3) = 0 implies alpha = frac12, 3 ### Step 2: Solve for x Since alpha = frac1sqrtx implies x = frac1alpha^2. For alpha = frac13 implies x = 9 For alpha = frac12 implies x = 4 For alpha = frac23 implies x = frac94 For alpha = 3 implies x = frac19 All 4 values are positive and valid. ### Pattern Recognition Always check constraints first (x > 0 due to sqrtx in denominator). Since all roots alpha > 0, every single algebraic root maps to a real distinct solution. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Quadratic Equations
Q9 jee_main_2024_01_february_morning Equations Reducible to Quadratic Form
Let S=\xin R:(sqrt3+sqrt2)^x+(sqrt3-sqrt2)^x=10\. Then the number of elements in S is:
  • A. 4
  • B. 0
  • C. 2
  • D. 1

Solution

### Related Formula Conjugate Surd Identity: (sqrta + sqrtb)(sqrta - sqrtb) = a - b ### Core Logic Observe the base components of the exponents: (sqrt3 + sqrt2)(sqrt3 - sqrt2) = 3 - 2 = 1 Therefore, we can express one base as the reciprocal of the other: sqrt3 - sqrt2 = frac1sqrt3 + sqrt2 The equation becomes: (sqrt3 + sqrt2)^x + frac1(sqrt3 + sqrt2)^x = 10 ### Step 1: Formulate the Quadratic Equation Let (sqrt3 + sqrt2)^x = t. Then: t + frac1t = 10 implies t^2 - 10t + 1 = 0 Solving for t using the quadratic formula: t = frac10 pm sqrt(-10)^2 - 4(1)(1)2 = frac10 pm sqrt962 = 5 pm 2sqrt6 ### Step 2: Solve for x Notice that (5 pm 2sqrt6) can be written as square powers of the original base: (sqrt3 pm sqrt2)^2 = 3 + 2 pm 2sqrt6 = 5 pm 2sqrt6 Thus, we have: - For t = 5 + 2sqrt6 implies (sqrt3 + sqrt2)^x = (sqrt3 + sqrt2)^2 implies x = 2 - For t = 5 - 2sqrt6 implies (sqrt3 + sqrt2)^x = (sqrt3 - sqrt2)^2 = (sqrt3 + sqrt2)^-2 implies x = -2 Therefore, the distinct real solutions are x = 2 and x = -2. The number of elements in set S is 2. ### Pattern Recognition Sees: Rational conjugate bases added with inverse matching variables. Shortcut: Whenever you see an equation of the form A^x + B^x = C where AB = 1, the solution will always be symmetric (pm x_0). Checking x=2 explicitly gives (sqrt3+sqrt2)^2 + (sqrt3-sqrt2)^2 = (5+2sqrt6) + (5-2sqrt6) = 10, confirming pm 2 immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Quadratic Equations Class 9 Mathematics: Number Systems (Rationalization)
Q28 jee_main_2024_30_january_evening Modulus Equations
The number of real solutions of the equation xleft(x^2 + 3|x| + 5|x - 1| + 6|x - 2|right) = 0 is
Numerical Answer. Answer: 1 to 1

Solution

### Related Formula textZero Product Property: A cdot B = 0 implies A = 0 text or B = 0 ### Core Logic Given equation: xleft(x^2 + 3|x| + 5|x - 1| + 6|x - 2|right) = 0 This factors into two possibilities: 1) x = 0 2) x^2 + 3|x| + 5|x - 1| + 6|x - 2| = 0 ### Step 1: Evaluating the Modulus Term Look at the second factor: f(x) = x^2 + 3|x| + 5|x - 1| + 6|x - 2|. Notice that all terms inside are strictly non-negative: - x^2 ge 0 - 3|x| ge 0 - 5|x - 1| ge 0 - 6|x - 2| ge 0 For the sum to be 0, every single term must be simultaneously zero. x^2 = 0 implies x = 0 However, if x = 0, then 5|x-1| = 5(1) = 5 neq 0. Therefore, there is no real value of x that makes this entire second factor equal to zero. ### Step 2: Conclusion The only valid solution to the equation is x = 0 from the first factor. Thus, there is exactly 1 real solution. ### Pattern Recognition A sum of absolute values and squares set to 0 requires all individual components to hit 0 concurrently. If they have different zero-nodes (0, 1, 2), the sum can never be 0. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Quadratic Equations
Q22 jee_main_2024_31_jan_evening Roots of Quadratic Equation
Let a, b, c be the length of three sides of a triangle satisfying the condition (a^2 + b^2)x^2 - 2b(a + c)x + (b^2 + c^2) = 0. If the set of all possible values of x is the interval (alpha, beta) then 12(alpha^2 + beta^2) is equal to
Numerical Answer. Answer: 36 to 36

Solution

### Core Logic Given equation: (a^2+b^2)x^2 - 2b(a+c)x + b^2+c^2 = 0. Expand and rearrange into perfect squares: (a^2x^2 - 2abx + b^2) + (b^2x^2 - 2bcx + c^2) = 0 (ax - b)^2 + (bx - c)^2 = 0 Since squares must be non-negative, each term is zero: ax - b = 0 implies x = fracba bx - c = 0 implies x = fraccb Thus, b = ax and c = bx = ax^2. Since a,b,c form a triangle, the triangle inequality holds: 1) a + b > c implies a + ax > ax^2 implies x^2 - x - 1 < 0 implies frac1-sqrt52 < x < frac1+sqrt52 2) a + c > b implies a + ax^2 > ax implies x^2 - x + 1 > 0 (Always true for real x) 3) b + c > a implies ax + ax^2 > a implies x^2 + x - 1 > 0 implies x > frac-1+sqrt52 or x < frac-1-sqrt52. Taking the intersection (and noting x > 0 since sides are positive): fracsqrt5-12 < x < fracsqrt5+12 So, alpha = fracsqrt5-12 and beta = fracsqrt5+12. Calculate 12(alpha^2 + beta^2): 12 left( frac6-2sqrt54 + frac6+2sqrt54 right) = 12 left( frac124 right) = 36 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations Class 11 Maths: Straight Lines
Q1 jee_main_2024_31_jan_morning Nature of Roots
For 0 < c < b < a, let (a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b) = 0 and alpha neq 1 be one of its root. Then, among the two statements (I) If alpha in (-1,0), then b cannot be the geometric mean of a and c (II) If alpha in (0,1), then b may be the geometric mean of a and c
  • A. textBoth (I) and (II) are true
  • B. textNeither (I) nor (II) is true
  • C. textOnly (II) is true
  • D. textOnly (I) is true

Solution

### Related Formula textSum of coefficients = 0 implies x = 1 text is a root. ### Core Logic Given f(x) = (a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b) = 0. Substituting x = 1: f(1) = a + b - 2c + b + c - 2a + c + a - 2b = 0 Thus, one root is 1. Let the other root be alpha. ### Step 1: Find the other root Product of roots = fracc + a - 2ba + b - 2c. Since one root is 1, we have: alpha cdot 1 = fracc + a - 2ba + b - 2c alpha = fracc + a - 2ba + b - 2c ### Step 2: Analyze Statement (I) If -1 < alpha < 0: -1 < fracc + a - 2ba + b - 2c < 0 This implies b > fraca + c2 and b + c < 2a. Therefore, b cannot be the Geometric Mean of a and c. Statement (I) is true. ### Step 3: Analyze Statement (II) If 0 < alpha < 1: 0 < fracc + a - 2ba + b - 2c < 1 This gives b > c and b < fraca + c2. Therefore, b may be the Geometric Mean between a and c. Statement (II) is true. ### Pattern Recognition When coefficients in a quadratic equation are cyclic and sum to 0, one root is always 1. The other root is directly c/a. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Quadratic Equations Class 11 Maths: Sequences and Series

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