Location of roots quadratic equation Free JEE Math tutorial

Location of Roots

Key Definitions

  1. Bounding Constant (k): A real number used as a reference point on the x-axis to determine the relative position of roots.
  2. Necessary and Sufficient Conditions: Inequalities involving the discriminant D, function value f(k), and vertex position that uniquely locate roots.
  3. Vertex of a Parabola: The point (-b/2a, -D/4a), representing the extremum of the quadratic function.

Theory and Concepts

Let f(x) = ax² + bx + c (assume a > 0). Let α and β be the roots (α ≤ β).

Case I: Both roots > k

  • D ≥ 0 (Real roots)
  • f(k) > 0
  • -b/2a > k (Vertex to the right of k)
Case II: Both roots < k

  • D ≥ 0
  • f(k) > 0
  • -b/2a < k (Vertex to the left of k)
Case III: k lies between the roots

  • f(k) < 0
  • Note: D > 0 is automatically satisfied if f(k) < 0 (when a > 0).
Case IV: Both roots lie between k₁ and k₂

  • D ≥ 0
  • f(k₁) > 0 and f(k₂) > 0
  • k₁ < -b/2a < k₂
Case V: Exactly one root in (k₁, k₂)
  • f(k₁) · f(k₂) < 0

Summary Table

Condition Geometric / Algebraic Criteria (a > 0)
Both roots > k D ≥ 0, -b/2a > k, f(k) > 0
Both roots < k D ≥ 0, -b/2a < k, f(k) > 0
k lies between roots f(k) < 0
Roots in (k₁, k₂) D ≥ 0, k₁ < -b/2a < k₂, f(k₁) > 0, f(k₂) > 0
Exactly one root in (k₁, k₂) f(k₁) · f(k₂) < 0

Solved Examples

Solved Example 1.1: Find ‘a’ such that both roots of x² – 6ax + 2 – 2a + 9a² = 0 are greater than 3.

Solution: Apply conditions for roots > 3:
1. D ≥ 0 ⇒ 36a² – 4(9a² – 2a + 2) ≥ 0 ⇒ 8a – 8 ≥ 0 ⇒ a ≥ 1.
2. f(3) > 0 ⇒ 9 – 18a + 9a² – 2a + 2 > 0 ⇒ 9a² – 20a + 11 > 0 ⇒ a ∈ (-∞, 1) ∪ (11/9, ∞).
3. Vertex -b/2a > 3 ⇒ 3a > 3 ⇒ a > 1.
Intersection: a ∈ (11/9, ∞).

Solved Example 1.2: Find m such that roots of x² – (m-3)x + m = 0 have opposite signs.

Solution: For opposite signs, 0 must be between roots.
Condition: f(0) < 0 ⇒ m < 0.
m ∈ (-∞, 0).

Solved Example 1.3: Find k such that one root of (k-5)x² – 2kx + k – 4 = 0 is < 1 and the other is > 2.

Solution: Points 1 and 2 lie between roots. Assuming (k-5) > 0:
f(1) < 0 ⇒ -9 < 0 (Always true).
f(2) < 0 ⇒ (k-24) < 0 ⇒ k < 24.
Intersection with leading coefficient constraint: k ∈ (5, 24).

Solved Example 1.4: Find m if roots of x² – mx + 1 = 0 are greater than 2.

Solution:
1. D ≥ 0 ⇒ m² – 4 ≥ 0 ⇒ m ≤ -2 or m ≥ 2.
2. f(2) > 0 ⇒ 5 – 2m > 0 ⇒ m < 2.5.
3. Vertex m/2 > 2 ⇒ m > 4.
Intersection of m < 2.5 and m > 4 is null. No such m exists.

Concept Application Exercise 1.3

  1. Find a such that roots of x² – 2ax + a² – a = 0 lie between -2 and 2.
  2. Find m such that both roots of x² – 6mx + 9m² – 2m + 2 = 0 are greater than 3.
  3. Find k if x² + (k-1)x + k + 2 = 0 has roots of opposite signs.
  4. Find a such that x² + 2(a-1)x + a + 5 = 0 has roots such that 4 lies between them.
  5. Find range of m for which one root of x² – 2mx + m² – 1 = 0 is < 0 and the other is > 2.
  6. Find values of p for which roots of x² – 2px + p² + p – 3 = 0 are real and less than 2.
  7. If both roots of x² – 2ax + a² – 1 = 0 lie in (-3, 5), find the range of a.
  8. Find a if roots of x² – 3x + a = 0 satisfy |α – β| = 1 and both are positive.
  9. Determine m so that (m-2)x² – 2(m-1)x + (m-3) = 0 has both roots positive.
  10. Find values of p for which both roots of x² – 2px + p² – 1 = 0 lie between -2 and 4.