Question 1170004
Let's analyze the given quadratic equation step-by-step:

**1. Existence of Roots:**

The given equation is:

mx² - 2(m - 1)x + m + 1 = 0

For the roots to exist, the discriminant (Δ) must be greater than or equal to zero.

Δ = b² - 4ac

where:
* a = m
* b = -2(m - 1)
* c = m + 1

Δ = [-2(m - 1)]² - 4(m)(m + 1)
Δ = 4(m² - 2m + 1) - 4m(m + 1)
Δ = 4m² - 8m + 4 - 4m² - 4m
Δ = -12m + 4

For roots to exist, Δ ≥ 0:
-12m + 4 ≥ 0
4 ≥ 12m
m ≤ 4/12
m ≤ 1/3

Therefore, the roots exist when m ≤ 1/3.

**2. Relation Independent of m:**

Let x1 and x2 be the roots of the equation.

Sum of roots: x1 + x2 = -b/a = 2(m - 1)/m
Product of roots: x1 * x2 = c/a = (m + 1)/m

Let's manipulate these equations to eliminate m.

From the sum of roots:
mx1 + mx2 = 2m - 2
m(x1 + x2 - 2) = -2
m = -2 / (x1 + x2 - 2)

From the product of roots:
mx1x2 = m + 1
m(x1x2 - 1) = 1
m = 1 / (x1x2 - 1)

Equating the two expressions for m:
-2 / (x1 + x2 - 2) = 1 / (x1x2 - 1)
-2(x1x2 - 1) = x1 + x2 - 2
-2x1x2 + 2 = x1 + x2 - 2
x1 + x2 + 2x1x2 - 4 = 0

This is the relation independent of m.

**3. Double Root (x1 = x2 = X):**

Substitute x1 = x2 = X into the relation:
X + X + 2X² - 4 = 0
2X + 2X² - 4 = 0
X² + X - 2 = 0
(X + 2)(X - 1) = 0
X = -2 or X = 1

**Case 1: X = -2**

Substitute X = -2 into the sum of roots:
2X = 2(m - 1)/m
2(-2) = 2(m - 1)/m
-4m = 2m - 2
-6m = -2
m = 1/3

**Case 2: X = 1**

Substitute X = 1 into the sum of roots:
2(1) = 2(m - 1)/m
2m = 2m - 2
0 = -2 (This is impossible)

Therefore, the only possible double root is X = -2, and the corresponding value of m is 1/3.

**Summary:**

1.  Roots exist when m ≤ 1/3.
2.  The relation independent of m is x1 + x2 + 2x1x2 - 4 = 0.
3.  The possible double root is X = -2, and the corresponding value of m is 1/3.