Directions (FOR NEXT TWO): Let `alpha` be the root of the equation `25cos^2theta + 5costheta - 12 = 0`, where `(pi)/(2) lt alpha lt pi`.
What is sin 2`alpha` equal to?

Options

A. `(24)/(25)`
B. `-(24)/(25)`
C. `-(5)/(12)`
D. `-(21)/(25)`

Answer

To solve the problem, we need to find the value of \( \sin 2\alpha \) given that \( \alpha \) is a root of the equation \( 25 \cos^2 \theta + 5 \cos \theta - 12 = 0 \) and \( \frac{\pi}{2} < \alpha < \pi \).

### Step-by-step Solution:

1. **Substitute \( \alpha \) into the equation**:
We replace \( \theta \) with \( \alpha \) in the equation:
\[ 25 \cos^2 \alpha + 5 \cos \alpha - 12 = 0 \]

2. **Let \( x = \cos \alpha \)**:
The equation becomes:
\[ 25x^2 + 5x - 12 = 0 \]

3. **Factor the quadratic equation**:
We can factor this equation. We look for two numbers that multiply to \( 25 \times -12 = -300 \) and add to \( 5 \). The numbers \( 20 \) and \( -15 \) work:
\[ 25x^2 + 20x - 15x - 12 = 0 \]
Grouping gives:
\[ 5x(5x + 4) - 3(5x + 4) = 0 \]
Factoring further:
\[ (5x - 3)(5x + 4) = 0 \]

4. **Find the values of \( x \)**:
Setting each factor to zero gives:
\[ 5x - 3 = 0 \quad \Rightarrow \quad x = \frac{3}{5} \]
\[ 5x + 4 = 0 \quad \Rightarrow \quad x = -\frac{4}{5} \]

5. **Determine the correct value of \( \cos \alpha \)**:
Since \( \alpha \) is in the second quadrant (\( \frac{\pi}{2} < \alpha < \pi \)), \( \cos \alpha \) must be negative. Therefore, we take:
\[ \cos \alpha = -\frac{4}{5} \]

6. **Calculate \( \sin \alpha \)**:
Using the Pythagorean identity:
\[ \sin^2 \alpha + \cos^2 \alpha = 1 \]
We have:
\[ \sin^2 \alpha + \left(-\frac{4}{5}\right)^2 = 1 \]
\[ \sin^2 \alpha + \frac{16}{25} = 1 \]
\[ \sin^2 \alpha = 1 - \frac{16}{25} = \frac{9}{25} \]
Thus:
\[ \sin \alpha = \sqrt{\frac{9}{25}} = \frac{3}{5} \]
Since \( \alpha \) is in the second quadrant, \( \sin \alpha \) is positive.

7. **Calculate \( \sin 2\alpha \)**:
Using the double angle formula:
\[ \sin 2\alpha = 2 \sin \alpha \cos \alpha \]
Substituting the values we found:
\[ \sin 2\alpha = 2 \cdot \frac{3}{5} \cdot \left(-\frac{4}{5}\right) = 2 \cdot \frac{3 \cdot -4}{25} = \frac{-24}{25} \]

### Final Answer:
\[ \sin 2\alpha = -\frac{24}{25} \]