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Mathematics

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.

Quadratic Equations

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Answer

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.

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