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Urgent! ! ! ! Several problems of quadratic mathematics in one variable.
1, according to Vieta's theorem:

x 1+x2=50 1

X 1+3= 125X2 = >x 1+x2= 126x2-3

So there are: 126x2-3=50 1.

x2=4

Bring it into the original equation: 16-50 1*4+k=0.

k= 1988

2、mx^2-2(m- 1)x-4=0

X 1=[2(m- 1)+ radical sign (4 (m-1) 2+16m)/2m = 2.

X2=[2(m- 1)- radical sign (4 (m-1) 2+16m)/2m = 2/m.

The above two formulas require discriminant > =0, that is, m & gt=- 1

When x=2. The equation is 4m-4 (m-1)-4 = 0m > =- 1

When x=2/m, the equation is: m= 1

To sum up, m= 1