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Quadratic Equations appeared 26 times across 3 years — 3% of Mathematics. This question is from Nature of Roots.

Year 2026 2025 2024 Total
Questions 11 10 5 26

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

Solution & Explanation

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

Reference Study Guides

More Quadratic Equations Previous-Year Questions — Page 3

Q8 jee_main_2026_28_january_morning Divisibility of Polynomials
Let S = x³ + ax² + bx + c : a, b, c in N and a, b, c ≤ 20 be a set of polynomials. Then the number of polynomials in S, which are divisible by x² + 2, is
  • A. 20
  • B. 6
  • C. 120
  • D. 10

Solution

Core Logic

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

x³ + ax² + bx + c = (x² + 2)(x + k)

where k is a constant.

Step 1: Equating Coefficients

Expand the right side:

(x² + 2)(x + k) = x³ + kx² + 2x + 2k

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

Step 2: Apply Constraints

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

Step 3: Final Count

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

Chapter Mix

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

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, (k)/(2)) from the line 3x + 4y + 5 = 0 is
  • A. 15
  • B. 5√(3)
  • C. 15√(5)
  • D. 12

Solution

Related Formula

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

D = b² - 4ac = 0

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

d = |Ax₀ + By₀ + C|√(A² + B²)
Core Logic

Let's expand the given equation:

(x² + 2x)(12 - k) = 2

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

λ x² + 2λ x - 2 = 0 (λ ≠ 0)
Step 1: Finding k

Set the discriminant to zero:

D = (2λ)² - 4(λ)(-2) = 0 4λ² + 8λ = 0 4λ(λ + 2) = 0

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

Thus:

12 - k = -2 k = 14
Step 2: Calculating perpendicular distance

The point of interest is:

(k, (k)/(2)) = (14, 7)

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

d = |3(14) + 4(7) + 5|√(3² + 4²) = (|42 + 28 + 5|)/(5) = (75)/(5) = 15
Pattern Recognition

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

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 R , for which both the roots of the equation x² - (p + 2)x + (2p + 9) = 0 are negative real numbers, be the interval (α, β] . Then β - 2α is equal to
  • A. 0
  • B. 9
  • C. 5
  • D. 20

Solution

Related Formula

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

  • D ≥ 0 (Real roots)
  • Sum of roots = -b/a < 0
  • Product of roots = c/a > 0
Core Logic

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

Condition 1: Discriminant D ≥ 0

D = [-(p + 2)]² - 4(1)(2p + 9) ≥ 0 p² + 4p + 4 - 8p - 36 ≥ 0 p² - 4p - 32 ≥ 0 (p - 8)(p + 4) ≥ 0 p in (-∞, -4] [8, ∞) (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

α + β = p + 2 < 0 p < -2 (ii)

Condition 3: Product of roots > 0

αβ = 2p + 9 > 0 p > -(9)/(2) (iii)
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)
  • Intersecting this with (i) limits the range cleanly to:
p in (-(9)/(2), -4]

Thus, α = -(9)/(2) and β = -4.

Step 3: Final Value Calculation

Calculate the requested target expression:

β - 2α = -4 - 2(-(9)/(2)) = -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.

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 x² - 2|x - 3| - 5 = 0, is
  • A. 26
  • B. 36
  • C. 30
  • D. 24

Solution

Related Formula
Sum of squares of roots = (α + β)² - 2αβ
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 (|x-2|+2)(|x-2|-1) = 0

Since |x-2| ≥ 0, we choose:

|x-2| = 1 x = 3 or 1

Sum of squares of roots = 3² + 1² = 10

Step 2: Evaluate Second Modulus Equation (Case Analysis)

For x² - 2|x - 3| - 5 = 0:

  • Case I (x ≥ 3): x² - 2x + 6 - 5 = 0 (x-1)² = 0 x = 1 (Rejected since x ≥ 3).
  • Case II (x < 3): x² + 2x - 6 - 5 = 0 x² + 2x - 11 = 0
Step 3: Final Combined Calculation

For the acceptable equation x² + 2x - 11 = 0, roots satisfy validation checks.

Sum of squares = (-2)² - 2(-11) = 4 + 22 = 26 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.

Chapter Mix

Class 11 Mathematics: Quadratic Equations

Q51 jee_main_2025_29_jan_evening Nature of Roots
If the set of all a in R, for which the equation 2x² + (a - 5)x + 15 = 3a has no real root, is the interval (α, β), and X = x in Z : α < x < β, then Σx in X x² is equal to
  • A. 2109
  • B. 2129
  • C. 2139
  • D. 2119

Solution

Related Formula

For a quadratic equation Ax² + Bx + C = 0 to have no real roots, its discriminant must be strictly negative:

D = B² - 4AC < 0
Core Logic

Rearranging the given equation into standard quadratic form:

2x² + (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 (a² - 10a + 25) - 8(15 - 3a) < 0 a² - 10a + 25 - 120 + 24a < 0 a² + 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 α = -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, , 0, 1, 2, 3, 4 Σx in X x² = (-18)² + (-17)² + + 4² = (1² + 2² + 3² + 4²) + (1² + 2² + + 18²) = (4 × 5 × 9)/(6) + (18 × 19 × 37)/(6) = 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 (n(n+1)(2n+1))/(6) efficiently.

Chapter Mix

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

More Quadratic Equations Questions — jee_main_2025_04_april_morning

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