Which of the following statements are correct
`E_(1)`) If a + b + c = 0 then 1 is a root of `ax^(2) + bx + c = 0`.
`E_(2)`) If `sin alpha, cos alpha` are the roots of the equation `ax^(2) + bx + c = 0` then `b^(2)-a^(2) =2ac`

Options

A. only `E_(1)`
B. only `E_(2)`
C. both `E_(1)` and `E_(2)`
D. neither `E_(2)` nor `E_(2)`

Answer

To determine the correctness of the statements \( E_1 \) and \( E_2 \), we will analyze each statement step by step.

### Step 1: Analyze Statement \( E_1 \)
**Statement**: If \( a + b + c = 0 \) then \( 1 \) is a root of \( ax^2 + bx + c = 0 \).

1. **Substituting \( x = 1 \)**:
\[ a(1)^2 + b(1) + c = 0 \]
This simplifies to:
\[ a + b + c = 0 \]
Since it is given that \( a + b + c = 0 \), we find that:
\[ 0 = 0 \]
Therefore, \( 1 \) is indeed a root of the equation.

### Conclusion for \( E_1 \):
**Result**: \( E_1 \) is correct.

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### Step 2: Analyze Statement \( E_2 \)
**Statement**: If \( \sin \alpha \) and \( \cos \alpha \) are the roots of the equation \( ax^2 + bx + c = 0 \), then \( b^2 - a^2 = 2ac \).

1. **Using Vieta's Formulas**:
- The sum of the roots \( \sin \alpha + \cos \alpha = -\frac{b}{a} \).
- The product of the roots \( \sin \alpha \cdot \cos \alpha = \frac{c}{a} \).

2. **Calculating \( (\sin \alpha + \cos \alpha)^2 \)**:
\[ (\sin \alpha + \cos \alpha)^2 = \sin^2 \alpha + \cos^2 \alpha + 2\sin \alpha \cos \alpha \]
Since \( \sin^2 \alpha + \cos^2 \alpha = 1 \), we have:
\[ 1 + 2\sin \alpha \cos \alpha = \left(-\frac{b}{a}\right)^2 \]
This gives us:
\[ 1 + 2\left(\frac{c}{a}\right) = \frac{b^2}{a^2} \]

3. **Rearranging the equation**:
\[ 1 + \frac{2c}{a} = \frac{b^2}{a^2} \]
Multiplying through by \( a^2 \):
\[ a^2 + 2ac = b^2 \]
Rearranging gives:
\[ b^2 - a^2 = 2ac \]

### Conclusion for \( E_2 \):
**Result**: \( E_2 \) is also correct.

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### Final Conclusion:
Both statements \( E_1 \) and \( E_2 \) are correct.

### Answer:
Both \( E_1 \) and \( E_2 \) are correct.

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