Mathematics
If a, b, c ∈ R, show that the roots of the equation (a - b)x2 + (b + c - a)x - c = 0 are rational.
Quadratic Equations
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Answer
(a - b)x2 + (b + c - a)x - c = 0, a ≠ b
Comparing (a - b)x2 + (b + c - a)x - c = 0 with ax2 + bx + c = 0 we get,
a = (a - b), b = (b + c - a) and c = -c.
We know that,
Discriminant (D) = b2 - 4ac
= (b + c - a)2 - 4 × (a - b) × (-c)
= [(b + c) - a]2 - 4 × (-ac + bc)
= [(b + c)2 + (a)2 - 2 × (b + c) × (a)] - (-4ac + 4bc)
= [(b)2 + (c)2 + 2 × b × c + a2 - 2 × (ab + ac)] + 4ac - 4bc
= (b2 + c2 + 2bc + a2 - 2ab - 2ac) + 4ac - 4bc
= a2 + b2 + c2 + 2bc - 2ab - 2ac + 4ac - 4bc
= a2 + b2 + c2 + 2bc - 4bc - 2ab - 2ac + 4ac
= a2 + b2 + c2 - 2bc - 2ab + 2ac
= a2 + c2 + 2ac + b2 - 2ab - 2bc
= a2 + c2 + 2ac + b2 - 2.(a + c).b
= (a + c - b)2
Thus D = (a + c - b)2, which is a perfect square.
The equation has rational roots. The roots are unequal if b ≠ a + c and equal if b = a + c (as then discriminant equals to zero).
Hence, (a - b)x2 + (b + c - a)x - c = 0 has rational roots. The roots are unequal if b ≠ a + c and equal if b = a + c.
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