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

Answer

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

### Step 1: Understand the Statement
The original statement can be expressed in the form of a conditional statement:
- Let \( p \): "x is a real number such that \( x^3 + 4x = 0 \)"
- Let \( q \): "x = 0"

Thus, the statement can be rewritten as:
- If \( p \) then \( q \) (i.e., \( p \implies q \))

### Step 2: Formulate the Contrapositive
The contrapositive of the statement \( p \implies q \) is:
- If not \( q \) (i.e., \( x \neq 0 \)), then not \( p \) (i.e., \( x^3 + 4x \neq 0 \))

This can be written as:
- If \( x \neq 0 \), then \( x^3 + 4x \neq 0 \)

### Step 3: Analyze the Case When \( x \neq 0 \)
Assume \( x \neq 0 \). We need to show that under this assumption, \( x^3 + 4x \neq 0 \).

### Step 4: Factor the Expression
We can factor the expression \( x^3 + 4x \):
\[ x^3 + 4x = x(x^2 + 4) \]

### Step 5: Analyze the Factors
1. The term \( x \) is not equal to zero by our assumption.
2. The term \( x^2 + 4 \) is always positive because:
- \( x^2 \) is non-negative for all real \( x \) (i.e., \( x^2 \geq 0 \)).
- Adding 4 to a non-negative number gives us a positive number (i.e., \( x^2 + 4 > 0 \)).

### Step 6: Conclude the Contrapositive
Since both factors \( x \) and \( x^2 + 4 \) are not zero when \( x \neq 0 \), we have:
\[ x(x^2 + 4) \neq 0 \]
Thus, \( x^3 + 4x \neq 0 \).

### Step 7: Final Conclusion
Since we have shown that if \( x \neq 0 \), then \( x^3 + 4x \neq 0 \), we conclude that the contrapositive is true. Therefore, the original statement "If \( x \) is a real number such that \( x^3 + 4x = 0 \), then \( x = 0 \)" is also true.

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