If roots of `ax^2+bx+c=0` are equal in magnitude but opposite in sign, then

Options

A. ` a lt 0 , c lt 0 `
B. `a lt 0 , c gt 0 , b gt 0 `
C. `a gt 0 , b =0 , c lt 0 `
D. `a gt 0 , b=0 , c gt 0 `

Answer

To solve the problem, we need to analyze the given quadratic equation \( ax^2 + bx + c = 0 \) under the condition that its roots are equal in magnitude but opposite in sign.

### Step-by-Step Solution:

1. **Understanding the Roots**:
- Let the roots of the equation be \( \alpha \) and \( -\alpha \). This means one root is \( \alpha \) and the other is \( -\alpha \).

2. **Sum of the Roots**:
- According to Vieta's formulas, the sum of the roots of the equation \( ax^2 + bx + c = 0 \) is given by:
\[ \text{Sum of roots} = -\frac{b}{a} \]
- For our roots \( \alpha \) and \( -\alpha \):
\[ \alpha + (-\alpha) = 0 \]
- Therefore, we have:
\[ 0 = -\frac{b}{a} \]
- This implies that:
\[ b = 0 \]

3. **Product of the Roots**:
- The product of the roots is given by:
\[ \text{Product of roots} = \frac{c}{a} \]
- For our roots:
\[ \alpha \cdot (-\alpha) = -\alpha^2 \]
- Thus, we have:
\[ -\alpha^2 = \frac{c}{a} \]
- This implies:
\[ c = -a\alpha^2 \]
- Since \( -\alpha^2 \) is always less than zero, we conclude that:
\[ \frac{c}{a} < 0 \]

4. **Analyzing the Signs**:
- For \( \frac{c}{a} < 0 \) to hold true, either:
- \( c > 0 \) and \( a < 0 \) or
- \( c < 0 \) and \( a > 0 \)

5. **Conclusion**:
- From the above analysis, we have established that:
- \( b = 0 \)
- The conditions on \( a \) and \( c \) must hold true as described.

### Final Answer:
The conditions that must be satisfied are:
- \( b = 0 \)
- Either \( c > 0 \) and \( a < 0 \) or \( c < 0 \) and \( a > 0 \).