How to do? Thanks.
更新1:
Let f(x) = x5 + 5qx3 + 5rx2 + t f(α) = α5 + 5qα3 + 5rα2 + t = 0 Case I: t = 0 x5 + 5qx3 + 5rx2 + t = 0 x5 + 5qx3 + 5rx2 = 0 x2(x3 + 5qx + 5r) = 0 …… (1)
更新2:
Suppose (1) has another non-zero double root b. Then x2(x – b)2(x – c) = 0 where c is not equal to b. x2(x2 – 2b + b2)(x – c) = 0 x2(x3 – cx2 + b2x – 2bx + 2bc – cb2) = 0 …… (2) Compare (1) with (2), we have c = 0, b2 – 2b = 5q, 2bc – cb2 = 5r
更新3:
So, r = 0 and therefore, f(x) = x5 + 5qx3 = x3(x2 + 5q), impossible to have another non-zero double root b! Thus, (1) implies that 0 is a double root of f(x). α = 0 Consider 3rα2 – 6q2α – 4qr + t = 3r(0)2 – 6q2(0) – 4qr + 0 = – 4qr = 0?
