If the difference of the roots of the equation `x^(2)-5x+6=0` is same as the difference of the roots of the equation `x^(2)+bx+c=0` then minimum value of `x^(2)+bx+c=0` is _________

Answer

To solve the problem, we need to find the minimum value of the quadratic equation \(x^2 + bx + c = 0\) given that the difference of the roots of this equation is the same as the difference of the roots of the equation \(x^2 - 5x + 6 = 0\).

### Step 1: Find the difference of the roots of the first equation
The first equation is:
\[ x^2 - 5x + 6 = 0 \]
For a quadratic equation \(ax^2 + bx + c = 0\), the difference of the roots can be calculated using the formula:
\[ |\alpha - \beta| = \frac{\sqrt{D}}{a} \]
where \(D = b^2 - 4ac\).

Here, \(a = 1\), \(b = -5\), and \(c = 6\).

Calculating \(D\):
\[ D = (-5)^2 - 4 \cdot 1 \cdot 6 = 25 - 24 = 1 \]

Now, substituting into the formula for the difference of the roots:
\[ |\alpha - \beta| = \frac{\sqrt{1}}{1} = 1 \]

### Step 2: Set up the second equation
The second equation is:
\[ x^2 + bx + c = 0 \]
For this equation, the difference of the roots is given by:
\[ |\alpha_1 - \beta_1| = \frac{\sqrt{D'}}{1} = \sqrt{D'} \]
where \(D' = b^2 - 4c\).

Since we know that the difference of the roots is the same as the first equation, we set:
\[ \sqrt{D'} = 1 \implies D' = 1 \]

This gives us:
\[ b^2 - 4c = 1 \]

### Step 3: Find the minimum value of the quadratic
The minimum value of the quadratic \(x^2 + bx + c\) occurs at:
\[ x = -\frac{b}{2a} = -\frac{b}{2} \]
Substituting this back into the quadratic to find the minimum value:
\[ \text{Minimum value} = \left(-\frac{b}{2}\right)^2 + b\left(-\frac{b}{2}\right) + c \]
\[ = \frac{b^2}{4} - \frac{b^2}{2} + c \]
\[ = \frac{b^2}{4} - \frac{2b^2}{4} + c = -\frac{b^2}{4} + c \]

### Step 4: Substitute \(c\) in terms of \(b\)
From the equation \(b^2 - 4c = 1\), we can express \(c\) as:
\[ c = \frac{b^2 - 1}{4} \]

Substituting this into the minimum value expression:
\[ \text{Minimum value} = -\frac{b^2}{4} + \frac{b^2 - 1}{4} \]
\[ = -\frac{b^2}{4} + \frac{b^2}{4} - \frac{1}{4} = -\frac{1}{4} \]

### Final Answer
Thus, the minimum value of \(x^2 + bx + c\) is:
\[ \boxed{-\frac{1}{4}} \]