If `b^(2) lt 2ac`, the equation `ax^(3)+bx^(2)+cx+d=0` has (where a, b, c, d in R and `a gt 0`)

Options

A. one real and two imaginary roots
B. all the roots real and distinct
C. all the roots real and equal
D. all the roots real and two of them equal different from the third

Answer

To solve the problem, we need to analyze the cubic equation \( ax^3 + bx^2 + cx + d = 0 \) under the condition that \( b^2 < 2ac \) and \( a > 0 \). We will determine the nature of the roots of this equation based on the given condition.

### Step-by-Step Solution:

1. **Understanding the Cubic Equation**:
The cubic equation is given by:
\[ ax^3 + bx^2 + cx + d = 0 \]
where \( a, b, c, d \in \mathbb{R} \) and \( a > 0 \).

2. **Using the Discriminant**:
The nature of the roots of a cubic equation can be analyzed using the discriminant. For a cubic equation, the discriminant \( D \) can be used to determine the nature of the roots:
- If \( D > 0 \), the equation has three distinct real roots.
- If \( D = 0 \), the equation has a multiple root (at least two roots are equal).
- If \( D < 0 \), the equation has one real root and two complex conjugate roots.

3. **Relating the Condition to the Discriminant**:
We are given that \( b^2 < 2ac \). We can relate this condition to the discriminant of the cubic equation. The discriminant \( D \) for a cubic equation can be expressed in terms of \( a, b, c, d \). However, we can also use the condition \( b^2 < 2ac \) to infer the nature of the roots.

4. **Assuming Three Distinct Real Roots**:
Let's assume that the cubic equation has three distinct real roots, denoted as \( \alpha, \beta, \gamma \). According to Vieta's formulas:
- \( \alpha + \beta + \gamma = -\frac{b}{a} \)
- \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \)
- \( \alpha\beta\gamma = -\frac{d}{a} \)

5. **Calculating the Sum of Squares of the Roots**:
We can calculate the sum of the squares of the roots:
\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \]
Substituting the values from Vieta's formulas:
\[ \alpha^2 + \beta^2 + \gamma^2 = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) \]
Simplifying this gives:
\[ \alpha^2 + \beta^2 + \gamma^2 = \frac{b^2}{a^2} - \frac{2c}{a} \]
\[ = \frac{b^2 - 2ac}{a^2} \]

6. **Analyzing the Result**:
Since \( \alpha^2, \beta^2, \gamma^2 \) are all non-negative (as they are squares), we have:
\[ \alpha^2 + \beta^2 + \gamma^2 \geq 0 \]
This implies:
\[ \frac{b^2 - 2ac}{a^2} \geq 0 \implies b^2 - 2ac \geq 0 \implies b^2 \geq 2ac \]

7. **Contradiction**:
However, we were given that \( b^2 < 2ac \). This leads to a contradiction since we cannot have both \( b^2 < 2ac \) and \( b^2 \geq 2ac \) simultaneously.

8. **Conclusion**:
Therefore, the assumption that the cubic equation has three distinct real roots must be false. The only possibility left is that the cubic equation has one real root and two complex conjugate roots.

### Final Answer:
The equation \( ax^3 + bx^2 + cx + d = 0 \) has one real root and two complex conjugate roots when \( b^2 < 2ac \).