Find the range of the real number m so that the equation x & # 178; + 2 (m-1) x + 2m + 6 = 0 (1) There are two real roots, one larger than 2 and the other smaller than 2 (2) There are two real roots, and both are larger than 1 (3) There are two real roots α and β, and 0 ＜ α ＜ 1 ＜ β is less than 4 (4) At least one positive root

Find the range of the real number m so that the equation x & # 178; + 2 (m-1) x + 2m + 6 = 0 (1) There are two real roots, one larger than 2 and the other smaller than 2 (2) There are two real roots, and both are larger than 1 (3) There are two real roots α and β, and 0 ＜ α ＜ 1 ＜ β is less than 4 (4) At least one positive root

Let f (x) = x & # 178; + 2 (m-1) x + 2m + 6, the symmetry axis of image parabola: x = 1-m, Δ = 4 (m-1) ^ 2-4 (2m + 6) = 4m ^ 2-16m-20 = 4 (m ^ 2-4m + 4) - 36 = 4 (m-2) ^ 2-36 ≥ 0, then m ≤ - 1 or m ≥ 5, f (2) 0, then f (2) = 4 + 4m-4 + 2m + 6 = 6m + 10-5 / 2, combined with Δ≥ 0, - 5 / 2 ≤ m ≤ - 1