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Quadratic Equations appeared 26 times across 3 years — 3% of Mathematics. This question is from Newton's Theorem on Roots.

Year 2026 2025 2024 Total
Questions 11 10 5 26

Let α and β be the roots of x² + √(3)x - 16 = 0, and γ and δ be the roots of x² + 3x - 1 = 0. If Pₙ = αⁿ + βⁿ and Qₙ = γⁿ + δⁿ, then P₂₅ + √(3)P₂₄2P₂₃ + Q₂₅ - Q₂₃Q₂₄ is equal to ______

Solution & Explanation

Related Formula

Newton's Theorem for roots: If α, β satisfy ax² + bx + c = 0 and Sₙ = αⁿ + βⁿ, then:

aSₙ + bSₙ₋₁ + cSₙ₋₂ = 0
Core Logic

Apply Newton's Theorem directly to the first equation x² + √(3)x - 16 = 0:

Pₙ + √(3)Pₙ₋₁ - 16Pₙ₋₂ = 0

For n = 25:

P₂₅ + √(3)P₂₄ - 16P₂₃ = 0 P₂₅ + √(3)P₂₄ = 16P₂₃

Dividing both sides by 2P₂₃:

P₂₅ + √(3)P₂₄2P₂₃ = 16P₂₃2P₂₃ = 8
Step 1: Evaluation of the Second Part

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

Qₙ + 3Qₙ₋₁ - Qₙ₋₂ = 0 Qₙ - Qₙ₋₂ = -3Qₙ₋₁

For n = 25:

Q₂₅ - Q₂₃ = -3Q₂₄

Dividing both sides by Q₂₄:

Q₂₅ - Q₂₃Q₂₄ = -3
Step 2: Total Calculation

Add both evaluated components:

Total Expression Value = 8 + (-3) = 5
Pattern Recognition

Shortcut: High sequential indices (25, 24, 23) indicate recurrence via Newton's Theorem. Relate Pₙ and Qₙ 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

More Quadratic Equations Previous-Year Questions — Page 4

Q jee_main_2025_28_jan_morning Equations Involving Modulus
The sum, of the squares of all the roots of the equation x² + |2x - 3| - 4 = 0, is:
  • A. 3(3 - √(2))
  • B. 6(3 - √(2))
  • C. 6(2 - √(2))
  • D. 3(2 - √(2))

Solution

Related Formula

Modulus definition rule:

|x| = cases x, & x ≥ 0 -x, & x < 0 cases
Core Logic

Analyze the roots by splitting into cases around the critical threshold x = (3)/(2):

Case I: x ≥ (3)/(2)

x² + 2x - 3 - 4 = 0 x² + 2x - 7 = 0 x = 2√(2) - 1

(We select the positive root since 2√(2)-1 ≥ 1.5).

Step 1: Evaluating the alternate domain branch

Case II: x < (3)/(2)

x² - (2x - 3) - 4 = 0 x² - 2x - 1 = 0 x = 1 - √(2)

(We select 1-√(2) since it satisfies the inequality constraint).

Step 2: Summing the Squares of the Roots
Sum of Squares = (2√(2) - 1)² + (1 - √(2))² = (8 - 4√(2) + 1) + (1 - 2√(2) + 2) = 12 - 6√(2) = 6(2 - √(2))
Pattern Recognition

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

Chapter Mix

Class 11 Maths: Quadratic Equations

Q68 jee_main_2025_04_april_morning Nature of Roots
Consider the equation x² + 4x - n = 0, where n in [20, 100] is a natural number. Then the number of all distinct values of n, for which the given equation has integral roots, is equal to
  • A. 7
  • B. 8
  • C. 6
  • D. 5

Solution

Related Formula

For quadratic equations with integer coefficients to have integral roots, the discriminant D = b² - 4ac must be a perfect square.

Core Logic

Rewrite using perfect square completing methods:

x² + 4x + 4 = n + 4 (x + 2)² = n + 4 x = -2 ± √(n + 4)

For x to be an integer, n + 4 must be a perfect square. Given range constraint 20 ≤ n ≤ 100:

24 ≤ n + 4 ≤ 104
Step 1: Identify Perfect Squares in Range

Find perfect squares between 24 and 104: 5² = 25 6² = 36 7² = 49 8² = 64 9² = 81 10² = 100

This gives exactly 6 distinct valid perfect squares.

Step 2: Conclusion

Thus, there are exactly 6 distinct integer values for n.

Pattern Recognition

Completing the square provides intuitive bounds quicker than running full discriminant inequalities. Match integer root sets directly to explicit numerical sequence counts.

Chapter Mix

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

Q67 jee_main_2025_07_april_evening Equations with Modulus
The number of real roots of the equation x | x - 2 | + 3 | x - 3 | + 1 = 0 is :
  • A. 4
  • B. 2
  • C. 1
  • D. 3

Solution

Related Formula

The definition of modulus function handles sub-intervals via critical points:

|x - a| = cases x - a & if x ≥ a -(x - a) & if x < a cases
Core Logic

The critical points are x = 2 and x = 3. We check the three distinct structural intervals:

Case I: x < 2

x(-(x - 2)) + 3(-(x - 3)) + 1 = 0 -x² + 2x - 3x + 9 + 1 = 0 x² + x - 10 = 0 x = -1 ± √(1 + 40)2 = -1 ± √(41)2

Checking domain constraint x < 2: -1 - √(41)2 ≈ (-1 - 6.4)/(2) = -3.7 < 2 (Valid root) -1 + √(41)2 ≈ (-1 + 6.4)/(2) = 2.7 < 2 (Rejected)

Step 1: Intermediate Interval Check

Case II: 2 ≤ x < 3

x(x - 2) + 3(-(x - 3)) + 1 = 0 x² - 2x - 3x + 9 + 1 = 0 x² - 5x + 10 = 0

Discriminant check: D = (-5)² - 4(1)(10) = 25 - 40 = -15 < 0. No real roots exist in this interval.

Step 2: Upper Interval Check

Case III: x ≥ 3

x(x - 2) + 3(x - 3) + 1 = 0 x² - 2x + 3x - 9 + 1 = 0 x² + x - 8 = 0 x = -1 ± √(1 + 32)2 = -1 ± √(33)2

Checking domain constraint x ≥ 3: -1 + √(33)2 ≈ (-1 + 5.74)/(2) = 2.37 < 3 (Rejected) -1 - √(33)2 < 0 (Rejected)

Thus, only 1 valid real root satisfies the conditional layout across all ranges.

Pattern Recognition

Always perform case-by-case boundaries checks on algebraic roots found inside absolute modulus problems to discard ghost solutions quickly.

Chapter Mix

Class 11 Mathematics: Quadratic Equations

Q67 jee_main_2025_24_jan_morning Roots of Advanced Polynomial Equations
The product of all the rational roots of the equation (x² - 9x + 11)² - (x - 4)(x - 5) = 3 is equal to :
  • A. 14
  • B. 7
  • C. 28
  • D. 41

Solution

Related Formula

For equations with repeating polynomial expressions, applying a variable substitution like t = P(x) reduces high-degree polynomials down to standard quadratics.

Core Logic

Expand the linear binomial product in the given equation equation:

(x-4)(x-5) = x² - 9x + 20

Rewrite the full expression in terms of the common variable pattern x² - 9x:

(x² - 9x + 11)² - (x² - 9x + 20) = 3

Let t = x² - 9x. Substituting this into the equation gives:

(t + 11)² - (t + 20) = 3 t² + 22t + 121 - t - 20 - 3 = 0 t² + 21t + 98 = 0
Step 1: Solve the Polynomial for Variable t

Factorize the quadratic expression:

(t + 14)(t + 7) = 0 t = -14 or t = -7
Step 2: Back-substitute and Isolate Rational Roots

Case 1: x² - 9x = -7 x² - 9x + 7 = 0 Check the discriminant value: D = (-9)² - 4(1)(7) = 81 - 28 = 53 (not a perfect square, so the roots are irrational).

Case 2: x² - 9x = -14 x² - 9x + 14 = 0 Factorize the quadratic expression:

(x - 7)(x - 2) = 0 x = 7 or x = 2

Both values are rational numbers.

Step 3: Calculate the Product of Rational Roots

Multiply the true rational roots together:

Product = 7 · 2 = 14
Pattern Recognition

Always check the discriminant D = b² - 4ac to filter out irrational radical components whenever the problem specifically asks for the product of rational roots only.

Chapter Mix

Class 11 Mathematics: Quadratic Equations

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)