Question:

Step 1: Analyze the discriminant of the first equation. We examine the equation $x^2 + 2\lambda x + \lambda^2 + 1 = 0$. The discriminant is: $$\Delta = (2\lambda)^2 - 4(1)(\lambda^2 + 1) = 4\lambda^2 - 4\lambda^2 - 4 = -4 < 0$$ Since the discriminant is negative, this equation has complex roots given by: $$x = \frac{-2\lambda \pm \sqrt{-4}}{2} = -\lambda \pm i$$ Step 2: Determine the nature of the common root. Since $ax^2 + bx + c = 0$ has real coefficients (as $a, b, c$ are side lengths of a triangle, hence real and positive), and it shares a common root with the first equation, the common root must be complex. For a polynomial with real coefficients, complex roots always occur in conjugate pairs. Therefore, if one root is $-\lambda + i$, the other must be $-\lambda - i$. This means both roots of $ax^2 + bx + c = 0$ are $-\lambda + i$ and $-\lambda - i$. Step 3: Express the second equation in terms of $\lambda$. We can write: $$ax^2 + bx + c = a(x - (-\lambda + i))(x - (-\lambda - i))$$ $$= a[(x + \lambda - i)(x + \lambda + i)]$$ $$= a[(x + \lambda)^2 - (i)^2]$$ $$= a[(x + \lambda)^2 + 1]$$ $$= a[x^2 + 2\lambda x + \lambda^2 + 1]$$ Comparing coefficients: $$b = 2a\lambda \quad \text{and} \quad c = a(\lambda^2 + 1)$$ Step 4: Apply the triangle inequality conditions. For $a, b, c$ to form sides of a triangle, all three triangle inequalities must hold: **Condition 1:** $a + b > c$ $$a + 2a\lambda > a(\lambda^2 + 1)$$ Dividing by $a > 0$: $$1 + 2\lambda > \lambda^2 + 1$$ $$2\lambda > \lambda^2$$ $$\lambda^2 - 2\lambda < 0$$ $$\lambda(\lambda - 2) < 0$$ This gives us: $0 < \lambda < 2$ **Condition 2:** $a + c > b$ $$a + a(\lambda^2 + 1) > 2a\lambda$$ Dividing by $a > 0$: $$1 + \lambda^2 + 1 > 2\lambda$$ $$\lambda^2 - 2\lambda + 2 > 0$$ $$(\lambda - 1)^2 + 1 > 0$$ This is always true for all real $\lambda$. **Condition 3:** $b + c > a$ $$2a\lambda + a(\lambda^2 + 1) > a$$ Dividing by $a > 0$: $$2\lambda + \lambda^2 + 1 > 1$$ $$\lambda^2 + 2\lambda > 0$$ $$\lambda(\lambda + 2) > 0$$ This gives us: $\lambda < -2$ or $\lambda > 0$ Step 5: Combine all conditions. From Condition 1: $0 < \lambda < 2$ From Condition 2: Always satisfied From Condition 3: $\lambda > 0$ (taking the relevant part since we already have $\lambda > 0$ from Condition 1) Combining all three conditions: $$0 < \lambda < 2$$ Therefore, the range of $\lambda$ is $(0, 2)$. The answer is **Option 1: $(0, 2)$**