It is known that when x = - 2, the value of the quadratic trinomial 2x2 + MX + 4 is equal to 18. When x is what, the value of the quadratic trinomial is 4?

It is known that when x = - 2, the value of the quadratic trinomial 2x2 + MX + 4 is equal to 18. When x is what, the value of the quadratic trinomial is 4?

∵ when x = - 2, the value of quadratic trinomial 2x2 + MX + 4 is equal to 18, ∵ 2x2 + MX + 4 = 18, the solution is m = - 3, substituting M = - 3 into 2x2 + MX + 4, we can get: 2x2-3x + 4, ∵ the value of quadratic trinomial is 4, ∵ 2x2-3x + 4 = 4, sorted out: 2x2-3x = 0, X (2x-3) = 0, then: x = 0, 2x-3 = 0, the solution is: X1 = 0

If we know that x is an integer and (x + 3) is 2 + (3-x) is 2 + (X & # 178; - 9) is (2x + 18), then the sum of the values of all qualified x is

Simplify the original formula = 2 / (x - 3), the problem should be out of the condition, such as it is a positive integer and so on

It is known that x is an integer, and (x + 3) 2 + (3-x) 2 + (X & sup2; - 9) 2x + 18

2/(x+3)+2/(3-x)+(2x+18)/(x^2-9)=2/(x+3)-2/(x-3)+(2x+18)/(x^2-9)=2(x-3)/(x^2-9)-2(x+3)/(x^2-9)+(2x+18)/(x^2-9)=(2x-6-2x-6+2x+18)/(x^2-9)=(2x+6)/(x^2-9)=2(x+3)/(x-3)(x+3)=2/(x-3).

If we know that x is an integer and 2x + 3 + 23 − x + 2x + 18x2 − 9 is an integer, then x is ()
A. 2 b. 3 C. 4 d. 5

The original formula = 2 (x − 3) − 2 (x + 3) + 2x + 18 (x + 3) (x − 3) = 2 (x + 3) (x + 3) (x − 3) = 2x − 3, when x-3 = 2, i.e. x = 5, the original formula value is an integer; when x-3 = 1, i.e. x = 4, the original formula value is an integer; when x-3 = - 1, i.e. x = 2, the original formula value is an integer; when x-3 = - 2, i.e. x = 1, the original formula value is an integer, then there are four X in the sign condition