It is known that the equation of X, where the square of x plus (2m minus 1) x plus the square of M is equal to 0, has two real roots x 1 and x 2

x1^2-x2^2=0
(x1+x2)(x1-x2)=0
X1 + x2 = 0 or x1-x2 = 0
x1+x2=0
Then, by the Weida theorem
x1+x2=-(2m-1)=0
m=1/2
The equation is x ^ 2 + 1 / 4 = 0
If there is no real solution, it does not hold
x1-x2=0
That is, the equation has two identical solutions
Then the discriminant is equal to 0
(2m-1)^2-4m^2=0
-4m+1=0
m=1/4
So m = 1 / 4

It is known that the equation of X, where the square of x plus (2m minus 1) x plus the square of M is equal to 0, has two real roots x 1 and x 2

It is known that the equation of X, where the square of x plus (2m minus 1) x plus the square of M is equal to 0, has two real roots x 1 and x 2

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