Question:

Step 1: Identify the roots and apply Vieta's formulas. Let the roots of the quadratic equation $x^2 + (3-\lambda)x + 2 = \lambda$ be $\alpha$ and $\beta$. Rewriting the equation in standard form: $$x^2 + (3-\lambda)x + (2-\lambda) = 0$$ By Vieta's formulas: - Sum of roots: $\alpha + \beta = -(3-\lambda) = \lambda - 3$ - Product of roots: $\alpha\beta = 2 - \lambda$ Step 2: Express the sum of squares of roots in terms of λ. We use the algebraic identity: $$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$$ Substituting the values from Step 1: $$\alpha^2 + \beta^2 = (\lambda - 3)^2 - 2(2 - \lambda)$$ Expanding: $$\alpha^2 + \beta^2 = \lambda^2 - 6\lambda + 9 - 4 + 2\lambda$$ $$\alpha^2 + \beta^2 = \lambda^2 - 4\lambda + 5$$ Step 3: Find the value of λ that minimizes the sum of squares. To find the minimum value, we take the derivative with respect to $\lambda$ and set it equal to zero: $$\frac{d}{d\lambda}(\lambda^2 - 4\lambda + 5) = 2\lambda - 4 = 0$$ Solving for $\lambda$: $$2\lambda = 4$$ $$\lambda = 2$$ Step 4: Verify this is a minimum. The second derivative is: $$\frac{d^2}{d\lambda^2}(\lambda^2 - 4\lambda + 5) = 2 > 0$$ Since the second derivative is positive, $\lambda = 2$ gives a minimum value. **Final Answer:** The value of $\lambda$ for which the sum of squares of the roots is least is $\lambda = 2$. This corresponds to **Option 2**.