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

Options

A. `a^(2)-b^(2) +2ac=0`
B. `(a+c)^(2) =b^(2) + c^(2)`
C. `b/a in | -sqrt2, sqrt2|`
D. `c/a in [ - 1/2 , 1/2 ] `

Answer

To solve the problem, we need to analyze the quadratic equation given the roots \( \sin \alpha \) and \( \cos \alpha \). The equation is:

\[ x^2 + bx + c = 0 \]

where \( c \neq 0 \).

### Step 1: Use Vieta's Formulas
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 \( r_1 \cdot r_2 = \frac{c}{a} \)

Here, the roots are \( \sin \alpha \) and \( \cos \alpha \). Therefore, we can write:

\[ \sin \alpha + \cos \alpha = -\frac{b}{a} \quad (1) \]
\[ \sin \alpha \cdot \cos \alpha = \frac{c}{a} \quad (2) \]

### Step 2: Square the Sum of the Roots
Now, we will square equation (1):

\[ (\sin \alpha + \cos \alpha)^2 = \left(-\frac{b}{a}\right)^2 \]

Expanding the left-hand 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 Product of Roots
From equation (2), we know \( \sin \alpha \cos \alpha = \frac{c}{a} \). Substituting this into our equation gives:

\[ 1 + 2 \left(\frac{c}{a}\right) = \frac{b^2}{a^2} \]

### Step 4: Rearranging the Equation
Rearranging this equation leads to:

\[ \frac{b^2}{a^2} = 1 + \frac{2c}{a} \]

Multiplying through by \( a^2 \):

\[ b^2 = a^2 + 2ac \]

### Step 5: Final Form
Rearranging gives us:

\[ a^2 - b^2 + 2ac = 0 \]

### Conclusion
Thus, we have derived the relationship:

\[ a^2 - b^2 + 2ac = 0 \]

This is the required result.

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