If a₁ < a₂ < a₃ < a₄ < a₅ < a₆, then the equation (x - a₁) (x - a₃) (x - a₅) + 3 (x - a₂) (x - a₄) (x - a₆) = 0, has

1) Three real roots

2) a root in (- ∞, a₁)

3) no real root in (a₁, a₂)

4) no real root in (a₅, a₆)

Solution: (1) Three real roots

Let f (x) = (x – a₁) (x – a₃) (x – a₅) + 3 (x – a₂) (x – a₄) (x – a₆)

As x approaches ∞, f (x) → ∞

f (a₁) = 3 (a₁ – a₂) (a₂ – a₄) (a₁ – a₆) < 0

Similarly f (a₂) > 0, f (a₃) > 0, f (a₄) < 0

f (a₅) < 0, f (a₆) > 0

f (x) changes sign in each interval (a₁ – a₂), (a₃ – a₄) and (a₅ – a₆)

Since f (x) = 0 is a cubic root of x, it will have only 1 root in each of the above sub intervals.