If a, b, c are the sides of a triangle having perimeter 2 than the roots of the equation `x^(2)-2sqrt(ab+bc+ca)(x)+abc+1=0` are

Options

A. real and distinct
B. may not be real
C. real but may not be distinct
D. real and equal

Answer

To solve the problem, we need to find the roots of the quadratic equation given by:

\[ x^2 - 2\sqrt{ab + bc + ca} \cdot x + (abc + 1) = 0 \]

where \( a, b, c \) are the sides of a triangle with a perimeter of 2.

### Step 1: Understand the Perimeter Condition
Given that the perimeter of the triangle is 2, we have:

\[ a + b + c = 2 \]

### Step 2: Use the Relationship of Sides in a Triangle
From the triangle inequality, we know:

1. \( a + b > c \)
2. \( b + c > a \)
3. \( c + a > b \)

These inequalities will help us ensure that \( a, b, c \) are valid sides of a triangle.

### Step 3: Calculate \( ab + bc + ca \)
Using the identity for the square of the sum:

\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \]

Substituting \( a + b + c = 2 \):

\[ 2^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \]

This simplifies to:

\[ 4 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \]

Rearranging gives:

\[ ab + bc + ca = \frac{4 - (a^2 + b^2 + c^2)}{2} \]

### Step 4: Calculate \( abc \)
The product \( abc \) can be expressed in terms of the sides of the triangle. However, without specific values for \( a, b, c \), we cannot simplify it further without loss of generality.

### Step 5: Calculate the Discriminant
The discriminant \( D \) of the quadratic equation is given by:

\[ D = b^2 - 4ac \]

For our equation, substituting \( a = 1 \), \( b = -2\sqrt{ab + bc + ca} \), and \( c = abc + 1 \):

\[ D = (-2\sqrt{ab + bc + ca})^2 - 4 \cdot 1 \cdot (abc + 1) \]

This simplifies to:

\[ D = 4(ab + bc + ca) - 4(abc + 1) \]

\[ D = 4(ab + bc + ca - abc - 1) \]

### Step 6: Analyze the Discriminant
For the roots to be real and distinct, we need \( D > 0 \):

\[ ab + bc + ca - abc - 1 > 0 \]

### Step 7: Conclude the Roots
From the analysis, we can conclude that the roots of the equation are real and distinct if the conditions on \( ab + bc + ca \) and \( abc \) hold true. Given that \( a, b, c \) are sides of a triangle, we can assert that the roots will indeed be real and distinct.

### Final Answer
Thus, the roots of the equation \( x^2 - 2\sqrt{ab + bc + ca} \cdot x + (abc + 1) = 0 \) are real and distinct.