Question

If a, b, c are rational, prove that the roots of the equation (b – c)x² + (c – a)x + (a – b) = 0 are also rational.

[Hint: D = [(c – a)² – 4(b – c)(a – b)] = (a + c – 2b)² ≥ 0.]

Answer 1

Comparing (b – c)x² + (c – a)x + (a – b) = 0 with ax² + bx + c = 0 we get,

a = (b – c), b = (c – a) and c = (a – b).

We know that,

Discriminant (D) = b² – 4ac

= (c – a)² – 4 × (b – c) × (a – b)

= [(c)² + (a)² – 2 × c × a] – 4 × (ba – b² – ac + bc)

= (c² + a² – 2ac) – (4ba – 4b² – 4ac + 4bc)

= c² + a² – 2ac – 4ba + 4b² + 4ac – 4bc

= c² + a² – 2ac + 4ac – 4ba + 4b² – 4bc

= c² + a² + 2ac – 4ba + 4b² – 4bc

= c² + a² + 2ac + (2b)² – 2.(c + a).2b

= (a + c – 2b)²

Thus, D = (a + c – 2b)², which is a perfect square.

The equation has rational roots.

Hence, proved that (b – c)x² + (c – a)x + (a – b) = 0 has rational roots.