Show that the statement .
p : 'If x is a real number such that `x^(3)+ 4x=0,` then x=0' is true by
Direct method

Answer

To show that the statement \( p: \) "If \( x \) is a real number such that \( x^3 + 4x = 0 \), then \( x = 0 \)" is true by direct method, we will follow these steps:

### Step 1: Start with the given equation
We start with the equation:
\[ x^3 + 4x = 0 \]

### Step 2: Factor the equation
We can factor out \( x \) from the left-hand side:
\[ x(x^2 + 4) = 0 \]

### Step 3: Set each factor to zero
For the product to be zero, at least one of the factors must be zero. Therefore, we have two cases to consider:
1. \( x = 0 \)
2. \( x^2 + 4 = 0 \)

### Step 4: Solve each case
- **Case 1**: From \( x = 0 \), we directly find that \( x = 0 \).
- **Case 2**: From \( x^2 + 4 = 0 \), we can rearrange it to:
\[ x^2 = -4 \]
Since \( -4 \) is negative, there are no real solutions for \( x^2 = -4 \). Thus, this case does not yield any real numbers.

### Step 5: Conclusion
The only solution that satisfies the original equation \( x^3 + 4x = 0 \) in the real number system is:
\[ x = 0 \]
Hence, we have shown that the statement \( p \) is true.