The positive integer m makes the quadratic equation of one variable with respect to x, M & # 178; X & # 178; - (2m-1) x + 1 = 0 have two real roots. Find the value of M. if it does not exist, explain the reason Students write as if it doesn't exist

The positive integer m makes the quadratic equation of one variable with respect to x, M & # 178; X & # 178; - (2m-1) x + 1 = 0 have two real roots. Find the value of M. if it does not exist, explain the reason Students write as if it doesn't exist

m^2x^2-(2m-1)x+1=0
The condition that there are two real roots is △ = [- (2m-1)] ^ 2-4m ^ 2 > 0
4m^2-4m+1-4m^2>0
4m

Let the roots of quadratic equation x2 + MX + n = 0 (m ≠ 0) be exactly m, N, then the value of Mn is

From Veda's theorem
m+n=-m mn=n
N = - 2m, m ≠ 0, so n ≠ 0
n(m-1)=0
N ≠ 0, so only m = 1, n = - 2m = - 2
mn=1×(-2)=-2

Two unequal real numbers m, n satisfy m ^ 2-6m = 24, n ^ 2-6n = 4, find the value of m ^ 2 + n ^ 2-4mn
I read it. They said that m and N are two of x ^ 2-6x-4 = 0. But I didn't understand it for a long time
It is m ^ 2-6m = 4.

Two unequal real numbers m, n satisfy M & # 178; - 6m = 4, n & # 178; - 6N = 4, find the value of M & # 178; + n & # 178; - 4Mn. From the meaning of the question, we can see that m and N are two of X ^ 2-6x-4 = 0, (substituting m and n into X & # 178; - 6x-4 = 0 respectively, that is M & # 178; - 6m = 4, n & # 178; - 6N = 4), so m + n = - (- 6) = 6, Mn = - 4m & # 178