Question:

Step 1: Define the roots and apply Vieta's formulas. Let the roots of the quadratic equation $x^2 + bx + c = 0$ be $\alpha$ and $\beta$. By Vieta's formulas: $$\alpha + \beta = -b$$ $$\alpha\beta = c$$ Step 2: Translate the given condition into equations. We are told that when 2 is subtracted from each root, the results are the reciprocals of the original roots. This gives us: $$\alpha - 2 = \frac{1}{\alpha}$$ $$\beta - 2 = \frac{1}{\beta}$$ Step 3: Simplify the condition for $\alpha$. Starting with $\alpha - 2 = \frac{1}{\alpha}$, multiply both sides by $\alpha$: $$\alpha^2 - 2\alpha = 1$$ $$\alpha^2 - 2\alpha - 1 = 0$$ Step 4: Simplify the condition for $\beta$. Similarly, from $\beta - 2 = \frac{1}{\beta}$: $$\beta^2 - 2\beta = 1$$ $$\beta^2 - 2\beta - 1 = 0$$ Step 5: Recognize that both roots satisfy the same equation. Since both $\alpha$ and $\beta$ satisfy the equation $x^2 - 2x - 1 = 0$, they must be the roots of this equation. Comparing $x^2 - 2x - 1 = 0$ with the original form $x^2 + bx + c = 0$: $$b = -2$$ $$c = -1$$ Step 6: Calculate $b^2 + c^2$. $$b^2 + c^2 = (-2)^2 + (-1)^2 = 4 + 1 = 5$$ **Final Answer:** The value of $b^2 + c^2$ is $\boxed{5}$, which corresponds to **Option 4**.