Question:

Step 1: Identify the condition for rational roots. For a quadratic equation to have only rational roots, its discriminant must be a perfect square (and non-negative). We need to find the discriminant of $nx^2 + (n+1)x + (n+1) = 0$. $$\Delta = (n+1)^2 - 4n(n+1)$$ Step 2: Simplify the discriminant. Factor out $(n+1)$ from the discriminant expression: $$\Delta = (n+1)^2 - 4n(n+1) = (n+1)[(n+1) - 4n] = (n+1)(1-3n)$$ Step 3: Determine when the discriminant is non-negative. For rational roots to exist, we need $\Delta \geq 0$: $$(n+1)(1-3n) \geq 0$$ This inequality holds when both factors have the same sign. Analyzing the sign: - $(n+1) = 0$ when $n = -1$ - $(1-3n) = 0$ when $n = \frac{1}{3}$ The product is non-negative when: $-1 \leq n \leq \frac{1}{3}$ Step 4: Identify integer values in the valid range. The integers satisfying $-1 \leq n \leq \frac{1}{3}$ are: $$n \in \{-1, 0\}$$ Step 5: Verify each integer value produces rational roots. **For $n = -1$:** The equation becomes: $-x^2 + 0 \cdot x + 0 = 0$, which simplifies to $-x^2 = 0$ This gives $x = 0$, which is rational. ✓ **For $n = 0$:** The equation becomes: $0 \cdot x^2 + x + 1 = 0$, which simplifies to $x + 1 = 0$ This gives $x = -1$, which is rational. ✓ Step 6: State the final answer. Both $n = -1$ and $n = 0$ satisfy the condition that the equation has only rational roots. The number of integers $n$ is **2**. **Answer: Option 3**