Let m be an integer and - 4 < m is less than 1. The equation x & # 178; - 2 (2m-3) x + 4m & # 178; - 14m + 8 = 0 has two unequal real roots Finding the value of M and the solution of the equation

Let m be an integer and - 4 < m is less than 1. The equation x & # 178; - 2 (2m-3) x + 4m & # 178; - 14m + 8 = 0 has two unequal real roots Finding the value of M and the solution of the equation

From - 4 to M0
So m > - 1 / 2
So m = 0
So x ^ 2 + 6x + 8 = 0
(x+2)(x+4)=0
x1=-2 x2=-4 m=0

Let the equation of line l be (M & sup2; - 2m-3) x + (2m & sup2; + m-1) y = 2m-6
Let the equation of the straight line l be (M & sup2; - 2m-3) x + (2m & sup2; + m-1) y = 2m-6. Under the following conditions, the value of M can not be obtained separately
1. The intercept on the x-axis is 1
2. The slope is 1
3. After fixed point P (- 1, - 1)
The format should also be standardized

The equation of straight line l can be reduced to: (M-3) (M + 1) x + (2m-1) (M + 1) y = 2 (M-3) 1. It is known that the straight line L passes through point (1,0), and the point is substituted into the equation to get (M-3) (M + 1) = 2 (M-3), and the solution is m = 3 or M = 12. The slope k = - (M-3) (M + 1) / (2m-1) (M + 1) = 1 (M is not equal to - 1 and 1 / 2), and the solution is m = 4 / 33

The equation (x ^ 2) + 2 (M + 3) x + 2m + 14 = 0 of X has two real roots, one of which is less than 1 and the other is more than 3

The discriminant is greater than 0
4(m+3)^2-4(2m+14)>0
m^2+4m-5>0
(m+5)(m-1)>0
m1
When x = 1 and 3, the value of the function is less than 0
When x = 1
x^2+2(m+3)x+2m+14=4m+21

The equation MX2 + 2 "m + 3" x + 2m + 14 = 0 has two real roots, one is greater than 4 and the other is less than 4, so we can find the value range of M
M can't be 0, the equation is as follows:
f(x)=x^2+2(1+3/m)x+2+14/m=0 ,
Because the opening of F (x) is upward, only f (4) is needed

You can just draw a picture, because if the opening is upward, if one is greater than 4 and the other is less than 4, there must be f (4)