If the roots of `ax^(2)+bx+c=0 " are " sin alpha and cos alpha " for some " alpha`, then which one of the following is correct ?

Options

A. ` a^(2) + b^(2) = 2ac`
B. ` b^(2) - c^(2) = 2ab`
C. ` b^(2) - a^(2) = 2 ac`
D. ` b^(2) + c^(2) = 2 ab `

Explanation

To solve the problem, we need to analyze the given quadratic equation \( ax^2 + bx + c = 0 \) and the roots \( \sin \alpha \) and \( \cos \alpha \).

### Step-by-Step Solution:

1. **Identify the roots**: The roots of the quadratic equation are given as \( \sin \alpha \) and \( \cos \alpha \).

2. **Sum of the roots**: According to Vieta's formulas, the sum of the roots can be expressed as:
\[ \sin \alpha + \cos \alpha = -\frac{b}{a} \]

3. **Product of the roots**: The product of the roots is given by:
\[ \sin \alpha \cdot \cos \alpha = \frac{c}{a} \]

4. **Use the identity for the sum of squares**: We know that:
\[ \sin^2 \alpha + \cos^2 \alpha = 1 \]

5. **Express the square of the sum of the roots**: We can square the sum of the roots:
\[ (\sin \alpha + \cos \alpha)^2 = \sin^2 \alpha + \cos^2 \alpha + 2 \sin \alpha \cos \alpha \]
Substituting the identity:
\[ (\sin \alpha + \cos \alpha)^2 = 1 + 2 \sin \alpha \cos \alpha \]

6. **Substituting the values from Vieta's formulas**: We can substitute the expressions for the sum and product of the roots:
\[ \left(-\frac{b}{a}\right)^2 = 1 + 2 \left(\frac{c}{a}\right) \]

7. **Simplifying the equation**: This gives us:
\[ \frac{b^2}{a^2} = 1 + \frac{2c}{a} \]
Multiplying through by \( a^2 \):
\[ b^2 = a^2 + 2ac \]

8. **Rearranging the equation**: We can rearrange this to find:
\[ b^2 - a^2 = 2ac \]

### Conclusion:
The correct relationship derived from the roots \( \sin \alpha \) and \( \cos \alpha \) in the quadratic equation \( ax^2 + bx + c = 0 \) is:
\[ b^2 - a^2 = 2ac \]