SOLUTION: Consider the equation of second degree in x : mx²-2(m-1)x+m+1=0 where m is a non zero real parameter 1)Following the values of m, find the existence of the roots of this equati

Question 1170004: Consider the equation of second degree in x :
mx²-2(m-1)x+m+1=0 where m is a non zero real parameter
1)Following the values of m, find the existence of the roots of this equation.
2)Find among the roots, when they exist, a relation independent of m.
3)Using this relation, find the possible double root (x1=x2=X)
Calculate the corresponding value of m.

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
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.

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