We know the quadratic equation of one variable about X (1): x2 + 2x + 2-m = 0. (1) if the equation has two unequal real roots, find the value range of real number m; (2) please use the conclusion of (1) to substitute any value of m into equation (1), and use the matching method to find the two real roots of the equation

We know the quadratic equation of one variable about X (1): x2 + 2x + 2-m = 0. (1) if the equation has two unequal real roots, find the value range of real number m; (2) please use the conclusion of (1) to substitute any value of m into equation (1), and use the matching method to find the two real roots of the equation

(1) ∵ the equation has two unequal real roots, namely △ ＞ 0, that is △ = b2-4ac = 4-4 (2-m) = 4m-4 ＞ 0, ∵ m ＞ 1; (2) for example, take M = 2 and substitute it into equation (1) to get x2 + 2x = 0, formula to get x2 + 2x + 12 = 12 (x + 1) 2 = 1x + 1 = ± 1 ∵ X1 = - 2, X2 = 0

1. It is known that the sign of the quadratic equation (M + 1) x square + X - (m-1) = m with respect to X is the same as that of the constant term
2. Given that the square of the equation AX + BX + C = 0 (a is not equal to 0) and another root is 1, find the value of a + B + C

1. The quadratic term is m + 1, and the primary term is - m-1-m. because of the same sign, the multiplication of (- 2m-1) by (M + 1) is greater than 0, so
2m ^ 2 + 3M + 1 x is less than 0, the opening of the function is upward, the two solutions of M are - 1 / 2 and - 1, that is, m belongs to - 1 / 2 to - 1
2. Substituting x = 1, the result is 0