Show that x = 2 is a root of the equation and solve it completely.

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

Taking the term (x – 2) common from R₂, we get

Applying R₃→ R₃ – R₁, we get

Taking the term (x + 3) common from R₃, we get

Applying C₁→ C₁ + C₃, we get

Expanding the determinant along C₁, we have

Δ = (x – 2)(x + 3)(x – 1)[(–3)(1) – (2)(1)]

⇒ Δ = (x – 2)(x + 3)(x – 1)(–5)

∴ Δ = –5(x – 2)(x + 3)(x – 1)

The given equation is Δ = 0.

⇒ –5(x – 2)(x + 3)(x – 1) = 0

⇒ (x – 2)(x + 3)(x – 1) = 0

Case – I:

x – 2 = 0 ⇒ x = 2

Case – II:

x + 2 = 0 ⇒ x = –3

Case – III:

x – 1 = 0 ⇒ x = 1

Thus, 2 is a root of the equation and its other roots are –3 and 1.