Let `f (x) =x ^(2) +bx+c,` minimum value of `f (x)` is -5, then abosolute value of the difference of the roots of `f (x)` is :

Options

A. 5
B. `sqrt20`
C. `sqrt15`
D. Can't be determined

Answer

To solve the problem step by step, we will analyze the given quadratic function and apply the necessary formulas.

### Step 1: Understand the quadratic function
The quadratic function is given by:
\[ f(x) = x^2 + bx + c \]
We know that the minimum value of this function is -5.

### Step 2: Identify the vertex of the parabola
The vertex of the parabola represented by the function \( f(x) \) occurs at:
\[ x = -\frac{b}{2a} \]
Since \( a = 1 \) (the coefficient of \( x^2 \)), we have:
\[ x = -\frac{b}{2} \]

### Step 3: Find the minimum value of the function
The minimum value of the function \( f(x) \) at the vertex is given by:
\[ f\left(-\frac{b}{2}\right) = -\frac{D}{4a} \]
where \( D \) is the discriminant of the quadratic equation. Since we know that the minimum value is -5, we can set up the equation:
\[ -\frac{D}{4} = -5 \]
This simplifies to:
\[ \frac{D}{4} = 5 \]
Thus, we find:
\[ D = 20 \]

### Step 4: Calculate the roots of the quadratic equation
The roots of the quadratic equation can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{D}}{2a} \]
Substituting \( a = 1 \) and \( D = 20 \):
\[ x = \frac{-b \pm \sqrt{20}}{2} \]

### Step 5: Determine the absolute value of the difference of the roots
Let the roots be \( x_1 \) and \( x_2 \):
- \( x_1 = \frac{-b - \sqrt{20}}{2} \)
- \( x_2 = \frac{-b + \sqrt{20}}{2} \)

The absolute value of the difference of the roots is given by:
\[ |x_2 - x_1| = \left| \frac{-b + \sqrt{20}}{2} - \frac{-b - \sqrt{20}}{2} \right| \]
This simplifies to:
\[ |x_2 - x_1| = \left| \frac{2\sqrt{20}}{2} \right| = |\sqrt{20}| \]

### Step 6: Final calculation
Since \( \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \), we have:
\[ |x_2 - x_1| = 2\sqrt{5} \]

### Conclusion
The absolute value of the difference of the roots of the function \( f(x) = x^2 + bx + c \) is:
\[ 2\sqrt{5} \]