When m is a value, the equation (M + 1) x ^ 2 - (2m-1) x + M-1 = 0 has real roots? (1) when m is a value, the equation always has two real roots? 2. When the equation has two real roots and the sum of squares of the two real roots is equal to 4, find the value of M

When m is a value, the equation (M + 1) x ^ 2 - (2m-1) x + M-1 = 0 has real roots? (1) when m is a value, the equation always has two real roots? 2. When the equation has two real roots and the sum of squares of the two real roots is equal to 4, find the value of M

On the equation of X (M + 1) x ^ 2 - (2m-1) x + M-1 = 0
(1)
When m + 1 = 0, i.e. M = - 1, the original equation is
3x-2 = 0, x = 2 / 3, there is only one real root, which is not in line with the meaning of the problem
When m + 1 ≠ 0, i.e. m ≠ - 1, the equation is quadratic
If there are two real roots
Δ=(2m-1)²-4(m+1)(m-1)≥0
That is, (4m & # 178; - 4m + 1) - 4 (M & # 178; - 1) ≥ 0
-4m+5≥0
M ≤ 5 / 4 and m ≠ - 1
When m ≤ 5 / 4 and m ≠ - 1, the equation always has two real roots
（2）
Let two roots of real numbers be X1 and x2. According to the Weida theorem
x1+x2=(2m-1)/(m+1) ,x1x2=(m-1)/(m+1)
∵x²1+x²2=4
∴（x1+x2)²-2x1x2=4
That is, (2m-1) &# 178; / (M + 1) &# 178; - 2 (m-1) / (M + 1) = 4
∴（2m-1)²-2(m-1)(m+1)=4(m+1)²
The results are as follows
4m²-4m+1-2m²+2=4m²+8m+4
2m²+12m+1=0
M = (- 6 - √ 34) / 2 or M = (- 6 + √ 34) / 2