Let

We need to find the roots of Δ = 0.

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₂→ R₂ – R₁, we get

Applying R₃→ R₃ – R₁, we get

Taking (a – x) and (b – x) common from R₂ and R₃, we get

Expanding the determinant along C₁, we have

Δ = (a – x)(b – x)(1)[(1)(b + x) – (1)(a + x)]

⇒ Δ = (a – x)(b – x)[b + x – a – x]

∴ Δ = (a – x)(b – x)(b – a)

The given equation is Δ = 0.

⇒ (a – x)(b – x)(b – a) = 0

However, a ≠ b according to the given condition.

⇒ (a – x)(b – x) = 0

Case – I:

a – x = 0 ⇒ x = a

Case – II:

b – x = 0 ⇒ x = b

Thus, a and b are the roots of the given determinant equation.