If `r_(1), r_(2), r_(3)` are radii of the escribed circles of a triangle ABC and r it the radius of its incircle, then the root(s) of the equation `x^(2)-r(r_(1)r_(2)+r_(2)r_(3)+r_(3)r_(1))x+(r_(1)r_(2)r_(3)-1)=0` is/are :

Options

A. `r_(1)`
B. `r_(2)+r_(3)`
C. 1
D. `r_(1)r_(2)r_(3)-1`

Answer

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

\[ x^2 - r(r_1 r_2 + r_2 r_3 + r_3 r_1)x + (r_1 r_2 r_3 - 1) = 0 \]

where \( r_1, r_2, r_3 \) are the radii of the escribed circles of triangle \( ABC \) and \( r \) is the radius of its incircle.

### Step 1: Identify the components of the equation
The equation is in the standard quadratic form \( ax^2 + bx + c = 0 \), where:
- \( a = 1 \)
- \( b = -r(r_1 r_2 + r_2 r_3 + r_3 r_1) \)
- \( c = r_1 r_2 r_3 - 1 \)

### Step 2: Use the quadratic formula
The roots of a quadratic equation can be found using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Substituting the values of \( a, b, \) and \( c \):

\[ x = \frac{r(r_1 r_2 + r_2 r_3 + r_3 r_1) \pm \sqrt{(r(r_1 r_2 + r_2 r_3 + r_3 r_1))^2 - 4(1)(r_1 r_2 r_3 - 1)}}{2} \]

### Step 3: Simplify the expression under the square root
Let’s denote \( P = r_1 r_2 r_3 \). Then we have:

\[ x = \frac{r(r_1 r_2 + r_2 r_3 + r_3 r_1) \pm \sqrt{(r(r_1 r_2 + r_2 r_3 + r_3 r_1))^2 - 4(P - 1)}}{2} \]

### Step 4: Check for specific roots
We can check if \( x = 1 \) is a root of the equation:

Substituting \( x = 1 \):

\[ 1^2 - r(r_1 r_2 + r_2 r_3 + r_3 r_1)(1) + (r_1 r_2 r_3 - 1) = 0 \]

This simplifies to:

\[ 1 - r(r_1 r_2 + r_2 r_3 + r_3 r_1) + (r_1 r_2 r_3 - 1) = 0 \]

Which can be rearranged to:

\[ r_1 r_2 r_3 - r(r_1 r_2 + r_2 r_3 + r_3 r_1) = 0 \]

This means that \( x = 1 \) is indeed a root.

### Step 5: Find the second root
Let the second root be denoted as \( \alpha \). The sum of the roots of the quadratic equation is given by:

\[ 1 + \alpha = r(r_1 r_2 + r_2 r_3 + r_3 r_1) \]

Thus, we can express \( \alpha \) as:

\[ \alpha = r(r_1 r_2 + r_2 r_3 + r_3 r_1) - 1 \]

### Step 6: Final roots
The roots of the quadratic equation are:

1. \( x = 1 \)
2. \( x = r(r_1 r_2 + r_2 r_3 + r_3 r_1) - 1 \)

### Conclusion
The roots of the given quadratic equation are:

\[ x = 1 \quad \text{and} \quad x = r_1 r_2 r_3 - 1 \]