On the univariate quadratic equation of X (m-1) x2 + (2m + 1) x + M2-1 = 0, then the value of M is 0______ .

On the univariate quadratic equation of X (m-1) x2 + (2m + 1) x + M2-1 = 0, then the value of M is 0______ .


Substituting x = 0 into the equation: M2-1 = 0, M-1 ≠ 0, the solution: M = - 1



When the value of M is in what range, the univariate quadratic equation mx2-4x + 5 = 0 and x2-4mx + 4m2-3m-3 = 0 have real roots. Is there an integer m such that the roots of the equation are also integers? If it exists, request; if it does not exist, please explain the reason


∵ the univariate quadratic equation mx2-4x + 5 = 0 and x2-4mx + 4m2-3m-3 = 0 have real roots, m ≠ 016 − 20m ≥ 016m2 − 4 (4m2 − 3M − 3) ≥ 0, the solution is - 1 ≤ m ≤ 45 and m ≠ 0, that is, when - 1 ≤ m ≤ 45 and m ≠ 0, the univariate quadratic equation mx2-4x + 5 = 0 and x2-4mx + 4m2-3m-3 = 0 have real roots 2-4x + 5 = 0 is - x2-4x + 5 = 0, the solution is X1 = - 5, X2 = 1, which is consistent with the meaning of the problem; when m = - 1, the equation x2-4mx + 4m2-3m-3 = 0 is x2 + 4x + 4 = 0, the solution is X1 = x2 = - 2, which is consistent with the meaning of the problem; therefore, there is an integer m = - 1, so that the root of the equation is also an integer



It is known that the focus of line L passing through line 3x + 4Y-2 = 0 and line 2x + y + 2 = 0 is p, and it is perpendicular to line x-2y-1 = 0
The equation for finding the line L
Find the area of L and axis besieged city!


1.
The intersection of 3x + 4Y-2 = 0 and 2x + y + 2 = 0 is p (- 2,2)
Because perpendicular to the line x-2y-1 = 0, the slope is - 2
The equation is Y-2 = - 2 (x + 2), that is, 2x + y + 2 = 0
two
The intersection of L and coordinate axis is (- 1,0), (0, - 2)
So the area is 1



The equations are solved by substitution elimination method and addition and subtraction elimination method respectively: x + y = 7, 3x + 5Y = 17
Two methods were used


X + y = 7, 1 3x + 5Y = 17, 2 2x + 3Y = 21, 2 - 3 2Y = - 4Y = - 2, substituting y = - 2 into 1 to get x = 9, so x = 9y = - 2x + y = 7, 1 3x + 5Y = 17, 2 from 1 to get y = 7-x, 3 from 1 to get 3x + 35-5x = 17-2x = - 18x = 9, substituting x = 9 into 3 to get y = - 2, so x = 9y