How to find common roots in Quadratic Equations free JEE Math Tutorial

Condition for Common Roots

Key Definitions

  1. Common Root: A value α is a common root if it satisfies two or more equations simultaneously.
  2. Resultant: The specific mathematical condition involving the coefficients of equations that must be true for a common root to exist.
  3. Simultaneous Equations: In this context, treating two quadratics as a system to solve for a single variable x.

Theory and Concepts

Consider two quadratic equations:

(i) a₁x² + b₁x + c₁ = 0 (a₁ ≠ 0)

(ii) a₂x² + b₂x + c₂ = 0 (a₂ ≠ 0)

1. Condition for Exactly One Common Root

Let α be the common root. By the rule of cross-multiplication:

α² / (b₁c₂ – b₂c₁) = α / (c₁a₂ – c₂a₁) = 1 / (a₁b₂ – a₂b₁)

Eliminating α, we get the Condition for one common root:

(c₁a₂ – c₂a₁)² = (a₁b₂ – a₂b₁)(b₁c₂ – b₂c₁)

2. Condition for Both Roots Common

If both roots are common, the equations are proportional:

a₁/a₂ = b₁/b₂ = c₁/c₂
Crucial Note: For equations with real coefficients, if one root is imaginary (p + iq), its conjugate (p – iq) must also be common. Thus, if equations share one imaginary root, they share both roots. The same logic applies to irrational roots with rational coefficients.

Solved Examples

Solved Example 1.1: Find k if x² – kx – 21 = 0 and x² – 3kx + 35 = 0 have a common root.

Solution: Let α be the root. Subtracting the equations: -2kα + 56 = 0 ⇒ kα = 28 ⇒ α = 28/k.
Substitute in eq 1: (28/k)² – k(28/k) – 21 = 0 ⇒ 784/k² – 49 = 0.
k² = 16 ⇒ k = ± 4.

Solved Example 1.2: If x² + ax + bc = 0 and x² + bx + ca = 0 have a common root, prove the other roots satisfy x² + cx + ab = 0.

Solution: Subtracting gives (a-b)α + c(b-a) = 0 ⇒ α = c.
If one root is c, product of roots bc (from eq 1) implies the other root is b. Similarly for eq 2, the other root is a.
Sum of roots: c + b = -a ⇒ a+b+c=0 ⇒ (a+b) = -c.
Equation with roots a, b: x² – (a+b)x + ab = 0 ⇒ x² + cx + ab = 0.

Solved Example 1.3: Find a if x² – 11x + a = 0 and x² – 14x + 2a = 0 have a common root.

Solution: Multiply first eq by 2 and subtract second eq: α² – 8α = 0.
If α = 0 ⇒ a = 0. If α = 8 ⇒ 64 – 88 + a = 0 ⇒ a = 24.
a ∈ {0, 24}.

Solved Example 1.4: If ax² + bx + c = 0 and x² + 2x + 3 = 0 have a common root and a, b, c ∈ ℝ, find a:b:c.

Solution: For x² + 2x + 3, D = 4 – 12 = -8 (imaginary roots).
Since coefficients are real, both roots must be common.
a/1 = b/2 = c/3 ⇒ a:b:c = 1:2:3.

Solved Example 1.5: Find k if x² – x – 6 = 0 and x² + kx – 2 = 0 have a common root.

Solution: Roots of x² – x – 6 are 3 and -2.
If α = 3: 9 + 3k – 2 = 0 ⇒ k = -7/3.
If α = -2: 4 – 2k – 2 = 0 ⇒ k = 1.
k ∈ {-7/3, 1}.

Concept Application Exercise 1.2

  1. Find a if 3x² + ax + 1 = 0 and 2x² + ax – 1 = 0 have a common root.
  2. If ax² + bx + c = 0 and bx² + cx + a = 0 have a common root (a,b,c ≠ 0), find (a³+b³+c³)/abc.
  3. If x² – cx + d = 0 and x² – ax + b = 0 have one common root and the second equation has equal roots, prove 2(b+d) = ac.
  4. If x² + kx + 64 = 0 and x² – 8x + k = 0 have a common root, find k.
  5. If x² + 3x + 5 = 0 and ax² + bx + c = 0 have a common root (a,b,c ∈ ℕ), find the minimum value of a + b + c.
  6. Find value of b for which x² + bx – 1 = 0 and x² + x + b = 0 have a common root.
  7. If one real root of 81x² + kx + 256 = 0 is the cube of the other root, find k.
  8. If x² + 2λx + 3λ = 0 and x² + (λ+1)x + (2λ+1) = 0 have a common root, find λ.
  9. If k(6x² + 3) + rx + 2x² – 1 = 0 and 6k(2x² + 1) + px + 4x² – 2 = 0 have both roots common, find 2r – p.
  10. Let α, β roots of x² – x + p = 0 and γ, δ roots of x² – 4x + q = 0. If α, β, γ, δ are in G.P., find integral p and q.