The equation `x^(2)+bx+c=0` has distinct roots. If `2` is subtracted from each root the result are the reciprocal of the original roots, then `b^(2)+c^(2)` is

Options

A. `2`
B. `3`
C. `4`
D. `5`

Explanation

To solve the equation \(x^2 + bx + c = 0\) given that if 2 is subtracted from each root, the results are the reciprocals of the original roots, we will follow these steps:

### Step 1: Define the Roots
Let the roots of the equation be \( \alpha \) and \( \beta \). According to Vieta's formulas:
- The sum of the roots:
\[ \alpha + \beta = -b \]
- The product of the roots:
\[ \alpha \beta = c \]

### Step 2: Set Up the Condition
According to the problem, if we subtract 2 from each root, we have:
\[ \alpha - 2 = \frac{1}{\alpha} \quad \text{and} \quad \beta - 2 = \frac{1}{\beta} \]

### Step 3: Rearrange the Equations
Rearranging the equations gives us:
\[ \alpha - 2 - \frac{1}{\alpha} = 0 \quad \Rightarrow \quad \alpha^2 - 2\alpha - 1 = 0 \]
\[ \beta - 2 - \frac{1}{\beta} = 0 \quad \Rightarrow \quad \beta^2 - 2\beta - 1 = 0 \]

### Step 4: Solve the Quadratic Equations
Now, we can solve these quadratic equations using the quadratic formula:
\[ \alpha = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \]
Thus, the roots are:
\[ \alpha = 1 + \sqrt{2}, \quad \beta = 1 - \sqrt{2} \]

### Step 5: Calculate \(b\) and \(c\)
Now we can find \(b\) and \(c\):
- \(b = -(\alpha + \beta) = -( (1 + \sqrt{2}) + (1 - \sqrt{2}) ) = -2\)
- \(c = \alpha \beta = (1 + \sqrt{2})(1 - \sqrt{2}) = 1 - 2 = -1\)

### Step 6: Calculate \(b^2 + c^2\)
Now we need to find \(b^2 + c^2\):
\[ b^2 = (-2)^2 = 4 \]
\[ c^2 = (-1)^2 = 1 \]
Thus,
\[ b^2 + c^2 = 4 + 1 = 5 \]

### Final Answer
The value of \(b^2 + c^2\) is \(5\).

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