For what value of k does `4x^2 + kx + 25 = 0` have exactly one real solution for x?`

Options

A. 5
B. 10
C. 15
D. 20

Answer

To determine the value of \( k \) for which the quadratic equation \( 4x^2 + kx + 25 = 0 \) has exactly one real solution, we need to analyze the condition for a quadratic equation to have a single solution.

### Step-by-step Solution:

1. **Identify the standard form of a quadratic equation**:
The standard form of a quadratic equation is given by:
\[ ax^2 + bx + c = 0 \]
In our case, we have:
\[ a = 4, \quad b = k, \quad c = 25 \]

2. **Condition for exactly one real solution**:
A quadratic equation has exactly one real solution when the discriminant \( D \) is equal to zero. The discriminant is given by:
\[ D = b^2 - 4ac \]
For our equation, substituting the values of \( a \), \( b \), and \( c \):
\[ D = k^2 - 4 \cdot 4 \cdot 25 \]

3. **Calculate the discriminant**:
\[ D = k^2 - 16 \cdot 25 \]
\[ D = k^2 - 400 \]

4. **Set the discriminant equal to zero**:
To find the value of \( k \) for which there is exactly one real solution, we set the discriminant to zero:
\[ k^2 - 400 = 0 \]

5. **Solve for \( k \)**:
Rearranging the equation gives:
\[ k^2 = 400 \]
Taking the square root of both sides results in:
\[ k = \pm 20 \]

6. **Conclusion**:
Therefore, the values of \( k \) for which the quadratic equation \( 4x^2 + kx + 25 = 0 \) has exactly one real solution are:
\[ k = 20 \quad \text{or} \quad k = -20 \]