Nature of Roots in Quadratic Equations Free JEE Math Tutorial

Nature of Roots

Key Definitions

  1. Discriminant (D): For ax² + bx + c = 0 (a ≠ 0), the quantity D = b² – 4ac is the discriminant. It determines the nature of roots without solving the equation.
  2. Real Roots: Values belonging to ℝ. Geometrically, these are where the parabola intersects or touches the x-axis.
  3. Imaginary (Complex) Roots: Involve the unit i = √-1. These occur when D < 0, meaning the parabola never meets the x-axis.
  4. Perfect Square Discriminant: If a, b, c are rational and D is a perfect square, the roots will be rational.

Theory and Concepts

For the quadratic equation ax² + bx + c = 0, the roots are calculated as:

x = (-b ± √D) / 2a

1. Classification Based on D

Case I: D > 0 (Real and Distinct Roots)

The parabola intersects the x-axis at two distinct points, α and β.

Case II: D = 0 (Real and Equal Roots)

The parabola touches the x-axis at exactly one point (α = β = -b/2a).

Case III: D < 0 (Imaginary Roots)

The parabola does not intersect the x-axis. Roots are complex conjugates (p ± iq).

2. Classification Based on Coefficients

Given a, b, c are Rational numbers (ℚ):

  • If D is a perfect square, roots are rational.
  • If D > 0 but not a perfect square, roots are irrational and occur in conjugate surds (p ± √q).

3. Integral Roots: If a = 1, b, c ∈ ℤ, and the roots are rational, then the roots must be integers.

Solved Examples

Solved Example 1.1: Find k for which x² – 2(1 + 3k)x + 7(3 + 2k) = 0 has equal roots.

Solution: Set D = 0.
[-2(1 + 3k)]² – 4(1)(7(3 + 2k)) = 0
4(1 + 9k² + 6k) – 28(3 + 2k) = 0
Divide by 4: 9k² – 8k – 20 = 0 ⇒ (9k + 10)(k – 2) = 0.
k = 2, -10/9

Solved Example 1.2: Find the nature of roots for 2x² – 5x + 3 = 0.

Solution: D = (-5)² – 4(2)(3) = 25 – 24 = 1.
Since D > 0 and a perfect square, roots are real, distinct, and rational.
x = (5 ± 1) / 4 ⇒ x = 3/2, 1.

Solved Example 1.3: Find k if kx(x – 2) + 6 = 0 has equal roots.

Solution: Standard form: kx² – 2kx + 6 = 0.
D = (-2k)² – 4(k)(6) = 4k² – 24k = 0.
4k(k – 6) = 0. Since a ≠ 0, k ≠ 0. Therefore, k = 6.

Solved Example 1.4: If one root of x² – 6x + k = 0 is 2 + √3, find k.

Solution: Irrational roots occur in pairs. Other root is 2 – √3.
Product of roots = k = (2 + √3)(2 – √3) = 4 – 3 = 1.
k = 1, Other root = 2 – √3.

Solved Example 1.5: Prove roots of (x-a)(x-b) = h² are always real.

Solution: x² – (a+b)x + ab – h² = 0.
D = (a+b)² – 4(ab – h²) = (a-b)² + 4h².
Since squares are non-negative, D ≥ 0. Roots are always real.

Solved Example 1.6: If 2 + i√3 is a root of x² + px + q = 0, find p and q.

Solution: Roots are 2 ± i√3.
Sum = -p = 4 ⇒ p = -4.
Product = q = 2² – (i√3)² = 4 + 3 = 7.
(p, q) = (-4, 7).

Solved Example 1.7: Find the value of a for which x² + ax – 4 = 0 has integral roots.

Solution: Product αβ = -4. Possible integer pairs: (1,-4), (-1,4), (2,-2).
Since -a = sum of roots:
For (1,-4), a = 3. For (-1,4), a = -3. For (2,-2), a = 0.
a ∈ {-3, 0, 3}.

Concept Application Exercise 1.1

  1. Find P such that 3x² – Px + 3 = 0 has equal roots.
  2. If roots of x² – 2cx + ab = 0 are real and unequal, prove roots of x² – 2(a+b)x + (a² + b² + 2c²) = 0 are imaginary.
  3. Find integral k for no real roots in (k-12)x² + 2(k-12)x + 2 = 0.
  4. If roots of (a²+b²)x² – 2(ac+bd)x + (c²+d²) = 0 are real, prove ad = bc.
  5. If x² + ax + b = 0 and x² + bx + a = 0 have a common root, show a + b + 1 = 0 or a = b.
  6. Find the range of a for both roots of x² – ax + 9 = 0 to be greater than 2.
  7. Find k such that x² + kx + k + 2 = 0 has equal roots.
  8. If the roots of x² – 5x + 16 = 0 are α, β and roots of x² + px + q = 0 are (α² + β²) and αβ/2, find p + q.
  9. Find values of a for which x² + ax + 9 = 0 has integral roots.
  10. Find the nature of roots for 3x² – 4√3x + 4 = 0.