If `sin alpha, cos alpha` are the roots of the equation `ax^(2)+bx+c=0`, then

Options

A. `a^(2)-b^(2)+2ac=0`
B. `(a-c)^(2)=b^(2)+c^(2)`
C. `a^(2)+b^(2)-2ac=0`
D. none of these

Answer

To solve the problem step by step, we will analyze the given information about the roots of the quadratic equation \( ax^2 + bx + c = 0 \) where the roots are \( \sin \alpha \) and \( \cos \alpha \).

### Step 1: Identify the sum and product of the roots
According to Vieta's formulas, for a quadratic equation \( ax^2 + bx + c = 0 \):
- The sum of the roots \( r_1 + r_2 = -\frac{b}{a} \)
- The product of the roots \( r_1 \cdot r_2 = \frac{c}{a} \)

In our case, the roots are \( \sin \alpha \) and \( \cos \alpha \). Therefore, we can write:
\[ \sin \alpha + \cos \alpha = -\frac{b}{a} \tag{1} \]
\[ \sin \alpha \cdot \cos \alpha = \frac{c}{a} \tag{2} \]

### Step 2: Square the sum of the roots
Next, we will square the sum of the roots from equation (1):
\[ (\sin \alpha + \cos \alpha)^2 = \left(-\frac{b}{a}\right)^2 \]
Expanding the left side:
\[ \sin^2 \alpha + \cos^2 \alpha + 2 \sin \alpha \cos \alpha = \frac{b^2}{a^2} \]
Using the Pythagorean identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \):
\[ 1 + 2 \sin \alpha \cos \alpha = \frac{b^2}{a^2} \]

### Step 3: Substitute the product of the roots
From equation (2), we know:
\[ \sin \alpha \cos \alpha = \frac{c}{a} \]
Substituting this into the equation we derived in Step 2:
\[ 1 + 2 \left(\frac{c}{a}\right) = \frac{b^2}{a^2} \]
Multiplying through by \( a^2 \) to eliminate the denominators:
\[ a^2 + 2ac = b^2 \]

### Step 4: Rearranging the equation
Rearranging the equation gives us:
\[ a^2 - b^2 + 2ac = 0 \]

### Conclusion
Thus, we have derived the relationship:
\[ a^2 - b^2 + 2ac = 0 \]