Key Concepts and Formulas

• Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
• Sum of Squares Identity: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.
• Vertex of a Parabola: The vertex of the parabola $ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$, which represents the minimum (if $a > 0$) or maximum (if $a < 0$) of the quadratic.

Step-by-Step Solution

1. Rewrite the Quadratic Equation

We begin by rewriting the given quadratic equation in the standard form $ax^2 + bx + c = 0$. Why this step? This is necessary to identify the coefficients $a$, $b$, and $c$ correctly for applying Vieta's formulas.

$x^2 + (3 - \lambda)x + 2 = \lambda$ Subtract $\lambda$ from both sides: $x^2 + (3 - \lambda)x + (2 - \lambda) = 0$ Now, we have $a = 1$, $b = 3 - \lambda$, and $c = 2 - \lambda$.

2. Apply Vieta's Formulas

Let $\alpha$ and $\beta$ be the roots of the quadratic equation. We use Vieta's formulas to express the sum and product of the roots in terms of $\lambda$. Why this step? Vieta's formulas provide a direct link between the roots and the coefficients of the polynomial, allowing us to express the sum and product of roots as functions of $\lambda$.

• Sum of roots: $\alpha + \beta = -\frac{b}{a} = -\frac{(3 - \lambda)}{1} = \lambda - 3$

• Product of roots: $\alpha\beta = \frac{c}{a} = \frac{2 - \lambda}{1} = 2 - \lambda$

3. Express the Sum of Squares in terms of $\lambda$

We want to find the value of $\lambda$ that minimizes $\alpha^2 + \beta^2$. We use the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$. Why this step? This identity allows us to express the sum of squares in terms of the sum and product of the roots, which we already know in terms of $\lambda$.

Substitute the expressions for $\alpha + \beta$ and $\alpha\beta$: $\alpha^2 + \beta^2 = (\lambda - 3)^2 - 2(2 - \lambda)$ Expand and simplify: $\alpha^2 + \beta^2 = (\lambda^2 - 6\lambda + 9) - (4 - 2\lambda)$ $\alpha^2 + \beta^2 = \lambda^2 - 6\lambda + 9 - 4 + 2\lambda$ $\alpha^2 + \beta^2 = \lambda^2 - 4\lambda + 5$ Let $S(\lambda) = \lambda^2 - 4\lambda + 5$.

4. Minimize the Quadratic Expression

We want to minimize $S(\lambda) = \lambda^2 - 4\lambda + 5$. This is a quadratic function of $\lambda$ with a positive leading coefficient, so it has a minimum value. We can find the value of $\lambda$ that minimizes $S(\lambda)$ using the vertex formula. Why this step? By expressing the quantity we want to minimize as a function of a single variable ($\lambda$), we can use standard techniques for finding the minimum of that function.

The $\lambda$-coordinate of the vertex is given by $\lambda = -\frac{b}{2a}$, where $a = 1$ and $b = -4$. $\lambda = -\frac{-4}{2(1)} = \frac{4}{2} = 2$ Therefore, the minimum value of $S(\lambda)$ occurs at $\lambda = 2$.

Common Mistakes & Tips

• Sign Errors: Pay close attention to signs, especially when applying Vieta's formulas and expanding expressions.
• Standard Form: Ensure the quadratic equation is in standard form $ax^2 + bx + c = 0$ before applying Vieta's formulas.
• Vertex Formula: Remember that the vertex formula gives the x-coordinate (in this case, the $\lambda$-coordinate) where the minimum or maximum occurs, not the minimum or maximum value itself.

Summary

By rewriting the quadratic equation in standard form and applying Vieta's formulas, we expressed the sum of squares of the roots as a quadratic function of $\lambda$. Minimizing this quadratic function using the vertex formula, we found that the least value of the sum of squares occurs when $\lambda = 2$.

Final Answer

The final answer is $\boxed{2}$, which corresponds to option (B).