f(x) is a polynomial of degree 3 with rational coefficients. If its graph touches the x-axis, show that it has three rational roots.
Solution
Without loss of generality, f(x) = x3 - ax2 + bx - c, where a, b, c are rational. Since the graph touches the x-axis, there is a repeated root, so we may take the roots to be h, h, k. Hence 2h + k = a, 2hk + k2 = b, h2k = c. Hence a2 - 3b = (h - k)2. Put r = ±√(a2 - 3b), where the sign is chosen so that h = a/3 + r/3, k = a/3 - 2r/3. We need to show that r is rational. If r is zero there is nothing to prove, so assume r is non-zero.
We have 9h2 = 2a2 - 3b + 2ar. Hence 27h2k = -2a3 + 9ab + (6b - 2a2)r. But 27h2k = 27c. So r = (27c + 2a3 - 9ab)/(2(3b - a2)). Note that 3b - 2a2 is non-zero because r is non-zero. So r is a rational combination of a, b, c and hence is rational.
© John Scholes
jscholes@kalva.demon.co.uk
1 July 2002