Prove that the product of the two real roots of x4 + x3 - 1 = 0 is a root of x6 + x4 + x3 - x2 - 1 = 0.
Solution
Let the roots of the quartic be a, b, c, d. We show that ab, ac, ad, bc, bd, cd are the roots of the sextic. We have a + b + c + d = -1, ab + ac + ad + bc + bd + cd = 0, 1/a + 1/b + 1/c + 1/d = 0, abcd = -1.
Let x6 + a5x5 + a4x4 + a3x3 + a2x2 + a1x + a0 be the sextic with roots ab, ac, ad, bc, bd, cd. Since their sum is zero, we have a5 = 0. Their product is (abcd)3, so a0 = -1.
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002