Proof: the equation x2 + MX + 1 = 0 of X has two negative real roots if and only if M ≥ 2

Proof: the equation x2 + MX + 1 = 0 of X has two negative real roots if and only if M ≥ 2

The following results are proved: (1) sufficiency: ∵ m ≥ 2, ∵ △ = M2-4 ≥ 0, the equation x2 + MX + 1 = 0 has real roots, let two of x2 + MX + 1 = 0 be x1, X2, according to Weida's theorem: x1x2 = 1 > 0, ∵ X1, X2 have the same sign, and ∵ X1 + x2 = - M ≤ - 2, ∵ x1, X2 are negative roots. (2) necessity: ∵ x2 + MX + 1 = 0 of two real roots x1, X2 are negative, and x1 · x2 = 1, ∵ m-2 = - (x1 + x2) - 2 = - (x1 + 1x1) - 2 = - X12 + 1 + 1x1 = - (x1 + 2x) +1) In conclusion, (1), (2) the proof of Zhiming problem