If all the roots of `x^3 + px + q = 0 p,q in R,q != 0` are real, then

Options

A. `p lt 0`
B. `p = 0`
C. `p gt 0`
D. `p gt q`

Explanation

To solve the problem, we need to analyze the cubic equation \( x^3 + px + q = 0 \) under the condition that all its roots are real.

### Step-by-Step Solution:

1. **Understanding the Cubic Function**:
We start with the cubic function \( f(x) = x^3 + px + q \). For this function to have three real roots, it must have a local maximum and a local minimum, which means the derivative must have two distinct real roots.

2. **Finding the Derivative**:
We calculate the derivative of \( f(x) \):
\[ f'(x) = 3x^2 + p \]

3. **Setting the Derivative to Zero**:
To find the critical points, we set the derivative equal to zero:
\[ 3x^2 + p = 0 \]
Rearranging gives:
\[ 3x^2 = -p \quad \Rightarrow \quad x^2 = -\frac{p}{3} \]

4. **Condition for Real Roots**:
For \( x^2 = -\frac{p}{3} \) to have real solutions, the right side must be non-negative:
\[ -\frac{p}{3} \geq 0 \quad \Rightarrow \quad p \leq 0 \]

5. **Conclusion**:
Since \( p \) must be less than or equal to zero for the cubic function to have three real roots, we conclude that:
\[ p < 0 \]
This condition ensures that the cubic function has the necessary shape (with a local maximum and minimum) to intersect the x-axis at three points.