For what values of k , the roots of the equation `x^(2) + 4x + k = 0` are real ?

Answer

To determine the values of \( k \) for which the roots of the equation \( x^2 + 4x + k = 0 \) are real, we need to analyze the discriminant of the quadratic equation. The discriminant \( D \) is given by the formula:

\[ D = b^2 - 4ac \]

where \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation \( ax^2 + bx + c = 0 \). In our case:

- \( a = 1 \)
- \( b = 4 \)
- \( c = k \)

### Step 1: Calculate the Discriminant

Substituting the values of \( a \), \( b \), and \( c \) into the discriminant formula:

\[ D = 4^2 - 4 \cdot 1 \cdot k \]

Calculating \( 4^2 \):

\[ D = 16 - 4k \]

### Step 2: Set Conditions for Real Roots

For the roots to be real, the discriminant must be greater than or equal to zero:

\[ D \geq 0 \]

This gives us the inequality:

\[ 16 - 4k \geq 0 \]

### Step 3: Solve the Inequality

Rearranging the inequality:

\[ 16 \geq 4k \]

Dividing both sides by 4:

\[ 4 \geq k \]

or equivalently,

\[ k \leq 4 \]

### Conclusion

The values of \( k \) for which the roots of the equation \( x^2 + 4x + k = 0 \) are real are:

\[ k \leq 4 \]