Two point `B(x_(1), y_(1))` and `C(x_(2), y_(2))` are such that `x_(1), x_(2)` are the roots of the equation `x^(2)+4x+3=0` and `y_(1),y_(2)` are the roots of the equation `y^(2)-y-6=0`. Also `x_(1)lt x_(2)` and `y_(1)gt y_(2)`. Coordinates of a third point A are (3, -5). If t is the length of the bisector of the angle BAC, then `l^(2)` is equal to.

Answer

To solve the problem step by step, we will first find the coordinates of points B and C, then calculate the lengths needed to find the length of the angle bisector of triangle ABC, and finally compute \( l^2 \).

### Step 1: Find the roots of the equations for points B and C.

1. **Finding \( x_1 \) and \( x_2 \)**:
The equation is \( x^2 + 4x + 3 = 0 \).
We can factor this as:
\[ (x + 3)(x + 1) = 0 \]
Thus, the roots are:
\[ x_1 = -3, \quad x_2 = -1 \]
Given that \( x_1 < x_2 \), we have:
\[ x_1 = -3, \quad x_2 = -1 \]

2. **Finding \( y_1 \) and \( y_2 \)**:
The equation is \( y^2 - y - 6 = 0 \).
We can factor this as:
\[ (y - 3)(y + 2) = 0 \]
Thus, the roots are:
\[ y_1 = 3, \quad y_2 = -2 \]
Given that \( y_1 > y_2 \), we have:
\[ y_1 = 3, \quad y_2 = -2 \]

### Step 2: Determine the coordinates of points B and C.

- Point B is \( B(x_1, y_1) = (-3, 3) \)
- Point C is \( C(x_2, y_2) = (-1, -2) \)

### Step 3: Identify the coordinates of point A.

Point A is given as \( A(3, -5) \).

### Step 4: Calculate the lengths of the sides of triangle ABC.

1. **Length \( BC \)**:
\[ BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{((-1) - (-3))^2 + ((-2) - 3)^2} \]
\[ = \sqrt{(2)^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29} \]

2. **Length \( CA \)**:
\[ CA = \sqrt{(x_A - x_C)^2 + (y_A - y_C)^2} = \sqrt{(3 - (-1))^2 + (-5 - (-2))^2} \]
\[ = \sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

3. **Length \( AB \)**:
\[ AB = \sqrt{(x_A - x_B)^2 + (y_A - y_B)^2} = \sqrt{(3 - (-3))^2 + (-5 - 3)^2} \]
\[ = \sqrt{(6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \]

### Step 5: Use the angle bisector theorem to find \( l^2 \).

The length of the angle bisector \( AD \) from A to BC can be calculated using the formula:
\[ l^2 = \frac{AB \cdot AC \cdot BC}{(AB + AC)^2} - \frac{BC^2}{4} \]

Substituting the lengths we found:
- \( AB = 10 \)
- \( AC = 5 \)
- \( BC = \sqrt{29} \)

Calculating \( l^2 \):
\[ l^2 = \frac{10 \cdot 5 \cdot \sqrt{29}}{(10 + 5)^2} - \frac{(\sqrt{29})^2}{4} \]
\[ = \frac{50\sqrt{29}}{225} - \frac{29}{4} \]
\[ = \frac{10\sqrt{29}}{45} - \frac{29}{4} \]

### Step 6: Simplifying the expression to find \( l^2 \).

To combine these fractions, we need a common denominator:
- The common denominator of 45 and 4 is 180.

Converting each term:
\[ \frac{10\sqrt{29}}{45} = \frac{40\sqrt{29}}{180} \]
\[ \frac{29}{4} = \frac{130.5}{180} \]

Thus,
\[ l^2 = \frac{40\sqrt{29}}{180} - \frac{130.5}{180} = \frac{40\sqrt{29} - 130.5}{180} \]

### Final Answer
Thus, \( l^2 \) is equal to \( \frac{40\sqrt{29} - 130.5}{180} \).

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