If the quadratic function y = - x ^ 2 + MX + M-3 has two different zeros, then the value range of M is The first floor is not △ 0, is it Please be more specific on the second floor

If the quadratic function y = - x ^ 2 + MX + M-3 has two different zeros, then the value range of M is The first floor is not △ 0, is it Please be more specific on the second floor

It is suggested that △ 0

If the two zeros of quadratic function f (x) = x2 + MX + (M + 4) are between 1 and 2, the value range of M is obtained

∵ the two zeros of quadratic function f (x) = x2 + MX + (M + 4) are between 1 and 2, ∵ f (− m2) ≤ 01 ≤ − M2 ≤ 2F (1) ≥ 0f (2) ≥ 0, that is (− m2) 2 − M22 + m + 4 ≤ 0 − 1 ≤ − M2 ≤ 21 + m + m + 4 ≥ 04 + 2m + m + 4 ≥ 0, the solution is: m ∈ [- 83, 4] ∵ the value range of M: [- 83, 4]

If the quadratic function y = x + MX + (M + 3) has two different zeros, what is the value range of M?

If the quadratic function y = x2 + MX + (M + 3) has two different zeros, then the value range of M is ()
A. （-2，6）B. [-2，6]C. {-2，6}D. （-∞，-2）∪（6，+∞）

∵ the quadratic function y = x2 + MX + (M + 3) has two different zeros, namely M2-4 (M + 3) > 0. The solution is m ∈ (- ∞, - 2) ∪ (6, + ∞), so D is chosen