To find the area of the triangle with vertices at \( (0,0) \), \( (41\alpha, 37) \), and \( (-37, 41\beta) \), where \( \alpha \) and \( \beta \) are the roots of the equation \( 3x^2 - 16x + 15 = 0 \), we will follow these steps:
### Step 1: Find the roots \( \alpha \) and \( \beta \)
The roots of the quadratic equation \( 3x^2 - 16x + 15 = 0 \) can be found using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 3 \), \( b = -16 \), and \( c = 15 \).
Calculating the discriminant:
\[
b^2 - 4ac = (-16)^2 - 4 \cdot 3 \cdot 15 = 256 - 180 = 76
\]
Now substituting back into the formula:
\[
x = \frac{16 \pm \sqrt{76}}{6}
\]
Simplifying \( \sqrt{76} = 2\sqrt{19} \):
\[
x = \frac{16 \pm 2\sqrt{19}}{6} = \frac{8 \pm \sqrt{19}}{3}
\]
Thus, the roots are:
\[
\alpha = \frac{8 - \sqrt{19}}{3}, \quad \beta = \frac{8 + \sqrt{19}}{3}
\]
### Step 2: Determine the coordinates of the triangle vertices
The vertices of the triangle are:
- \( A(0, 0) \)
- \( B(41\alpha, 37) = \left( 41 \cdot \frac{8 - \sqrt{19}}{3}, 37 \right) = \left( \frac{328 - 41\sqrt{19}}{3}, 37 \right) \)
- \( C(-37, 41\beta) = \left( -37, 41 \cdot \frac{8 + \sqrt{19}}{3} \right) = \left( -37, \frac{328 + 41\sqrt{19}}{3} \right) \)
### Step 3: Calculate the area of the triangle
The area \( A \) of a triangle given vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the formula:
\[
A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the coordinates:
\[
A = \frac{1}{2} \left| 0(37 - \frac{328 + 41\sqrt{19}}{3}) + \frac{328 - 41\sqrt{19}}{3} \left( \frac{328 + 41\sqrt{19}}{3} - 0 \right) + (-37)(0 - 37) \right|
\]
This simplifies to:
\[
A = \frac{1}{2} \left| \frac{328 - 41\sqrt{19}}{3} \cdot \frac{328 + 41\sqrt{19}}{3} + 37^2 \right|
\]
Calculating \( \frac{328 - 41\sqrt{19}}{3} \cdot \frac{328 + 41\sqrt{19}}{3} \):
This is a difference of squares:
\[
= \frac{(328)^2 - (41\sqrt{19})^2}{9} = \frac{107584 - 1681 \cdot 41}{9} = \frac{107584 - 68921}{9} = \frac{38663}{9}
\]
Now substituting back into the area formula:
\[
A = \frac{1}{2} \left| \frac{38663}{9} + 1369 \right| = \frac{1}{2} \left| \frac{38663 + 12321}{9} \right| = \frac{1}{2} \left| \frac{50984}{9} \right| = \frac{50984}{18}
\]
Calculating this gives:
\[
A \approx 2821.33
\]
### Final Answer
The area of the triangle is approximately \( 2821.33 \).
---
