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Let P(x) = x2 + ax + b be a quadratic polynomial with real coeffcients. Suppose there are real numbers s ≠ t such that P(s) = t and P(t) = s. Prove that b - st is a root of the equation x2 + ax + b - st = 0. |
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Answer» We have s2 + as + b = t; t2 + at + b = s: This gives (s2 - t2) + a(s - t) = (t - s). Since s ≠ t, we obtain s + t + a = -1. Adding the equations, we obtain s2 + t2 + a(s + t) + 2b = (s + t). Therefore (s + t)2 - 2st + a(s + t) + 2b = (s + t). Using s + t = -(1 + a), we obtain (1 + a)2 - 2st - a(1 + a) + 2b = -1 - a. Simplication gives st = 1 + a + b = P(1). This shows that x = 1 is a root of x2+ ax + b - st = 0. Since the product of roots is b - st, the other root is b - st. |
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