`cosα` is a root of the equation `25x^2+5x−12=0,−1 lt x lt 0`, then find the value of `sin2α` is:

Options

A. `24/25
B. `-12/25`
C. `-24/25`
D. `20/25`

Explanation

To solve the problem, we need to find the value of \( \sin 2\alpha \) given that \( \cos \alpha \) is a root of the equation \( 25x^2 + 5x - 12 = 0 \) and that \( -1 < x < 0 \).

### Step-by-Step Solution:

1. **Identify the quadratic equation**:
\[ 25x^2 + 5x - 12 = 0 \]

2. **Use the quadratic formula**:
The roots of the quadratic equation \( ax^2 + bx + c = 0 \) can be found using the formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 25 \), \( b = 5 \), and \( c = -12 \).

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 5^2 - 4 \cdot 25 \cdot (-12) = 25 + 1200 = 1225 \]

4. **Find the square root of the discriminant**:
\[ \sqrt{1225} = 35 \]

5. **Substitute into the quadratic formula**:
\[ x = \frac{-5 \pm 35}{50} \]

6. **Calculate the two possible roots**:
- First root:
\[ x_1 = \frac{-5 + 35}{50} = \frac{30}{50} = \frac{3}{5} \]
- Second root:
\[ x_2 = \frac{-5 - 35}{50} = \frac{-40}{50} = -\frac{4}{5} \]

7. **Choose the root that satisfies the condition \( -1 < x < 0 \)**:
The valid root is:
\[ \cos \alpha = -\frac{4}{5} \]

8. **Find \( \sin \alpha \)** using the Pythagorean identity:
\[ \sin^2 \alpha + \cos^2 \alpha = 1 \]
\[ \sin^2 \alpha + \left(-\frac{4}{5}\right)^2 = 1 \]
\[ \sin^2 \alpha + \frac{16}{25} = 1 \]
\[ \sin^2 \alpha = 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25} \]
\[ \sin \alpha = \pm \frac{3}{5} \]
Since \( \cos \alpha \) is negative, \( \alpha \) is in the second quadrant, thus:
\[ \sin \alpha = \frac{3}{5} \]

9. **Calculate \( \sin 2\alpha \)** using the double angle formula:
\[ \sin 2\alpha = 2 \sin \alpha \cos \alpha \]
\[ \sin 2\alpha = 2 \cdot \frac{3}{5} \cdot \left(-\frac{4}{5}\right) \]
\[ \sin 2\alpha = 2 \cdot \frac{3 \cdot -4}{5 \cdot 5} = \frac{-24}{25} \]

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