Key Concepts and Formulas

• Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$ with roots $r_1$ and $r_2$, we have $r_1 + r_2 = -\frac{b}{a}$ and $r_1 r_2 = \frac{c}{a}$.
• Algebraic Manipulation: Using substitution and simplification to solve for unknowns.
• Rationalization: Multiplying the numerator and denominator of a fraction by the conjugate to eliminate radicals in the denominator.

Step-by-Step Solution

Step 1: Apply Vieta's Formulas to the First Quadratic Equation

We have the equation $x^2 - 4\lambda x + 5 = 0$ with roots $\alpha$ and $\beta$. Applying Vieta's formulas:

• Sum of roots: $\alpha + \beta = -\frac{-4\lambda}{1} = 4\lambda$ (Equation 1)
• Product of roots: $\alpha \beta = \frac{5}{1} = 5$ (Equation 2)

Step 2: Apply Vieta's Formulas to the Second Quadratic Equation

We have the equation $x^2 - (3\sqrt{2} + 2\sqrt{3})x + 7 + 3\lambda \sqrt{3} = 0$ with roots $\alpha$ and $\gamma$. Applying Vieta's formulas:

• Sum of roots: $\alpha + \gamma = -\frac{-(3\sqrt{2} + 2\sqrt{3})}{1} = 3\sqrt{2} + 2\sqrt{3}$ (Equation 3)
• Product of roots: $\alpha \gamma = \frac{7 + 3\lambda \sqrt{3}}{1} = 7 + 3\lambda \sqrt{3}$ (Equation 4)

Step 3: Use the Given Condition to Express $\alpha$ in Terms of $\lambda$

We are given that $\beta + \gamma = 3\sqrt{2}$. We can express $\beta$ and $\gamma$ in terms of $\alpha$ and $\lambda$ using Equations 1 and 3:

• From Equation 1: $\beta = 4\lambda - \alpha$
• From Equation 3: $\gamma = 3\sqrt{2} + 2\sqrt{3} - \alpha$

Substitute these expressions into the given condition $\beta + \gamma = 3\sqrt{2}$: $(4\lambda - \alpha) + (3\sqrt{2} + 2\sqrt{3} - \alpha) = 3\sqrt{2}$ $4\lambda - \alpha + 3\sqrt{2} + 2\sqrt{3} - \alpha = 3\sqrt{2}$ $4\lambda + 2\sqrt{3} - 2\alpha = 0$ $2\alpha = 4\lambda + 2\sqrt{3}$ $\alpha = 2\lambda + \sqrt{3}$ (Equation 5)

Step 4: Substitute $\alpha$ into the Product of Roots Equation (Equation 2) to Find $\beta$ in Terms of $\lambda$

Substitute $\alpha = 2\lambda + \sqrt{3}$ into $\alpha \beta = 5$: $(2\lambda + \sqrt{3})\beta = 5$ $\beta = \frac{5}{2\lambda + \sqrt{3}}$ (Equation 6)

Step 5: Substitute $\alpha$ into the Product of Roots Equation (Equation 4) to Find $\gamma$ in Terms of $\lambda$

Substitute $\alpha = 2\lambda + \sqrt{3}$ into $\alpha \gamma = 7 + 3\lambda \sqrt{3}$: $(2\lambda + \sqrt{3})\gamma = 7 + 3\lambda \sqrt{3}$ $\gamma = \frac{7 + 3\lambda \sqrt{3}}{2\lambda + \sqrt{3}}$ (Equation 7)

Step 6: Substitute $\beta$ and $\gamma$ into the Given Condition $\beta + \gamma = 3\sqrt{2}$ and Solve for $\lambda$

Substitute Equations 6 and 7 into $\beta + \gamma = 3\sqrt{2}$: $\frac{5}{2\lambda + \sqrt{3}} + \frac{7 + 3\lambda \sqrt{3}}{2\lambda + \sqrt{3}} = 3\sqrt{2}$ $\frac{12 + 3\lambda \sqrt{3}}{2\lambda + \sqrt{3}} = 3\sqrt{2}$ $\frac{3(4 + \lambda \sqrt{3})}{2\lambda + \sqrt{3}} = 3\sqrt{2}$ $\frac{4 + \lambda \sqrt{3}}{2\lambda + \sqrt{3}} = \sqrt{2}$ $4 + \lambda \sqrt{3} = \sqrt{2}(2\lambda + \sqrt{3})$ $4 + \lambda \sqrt{3} = 2\lambda \sqrt{2} + \sqrt{6}$ $4 - \sqrt{6} = 2\lambda \sqrt{2} - \lambda \sqrt{3}$ $4 - \sqrt{6} = \lambda (2\sqrt{2} - \sqrt{3})$ $\lambda = \frac{4 - \sqrt{6}}{2\sqrt{2} - \sqrt{3}}$

Step 7: Rationalize the Denominator to Simplify $\lambda$

Multiply the numerator and denominator by the conjugate of the denominator, $2\sqrt{2} + \sqrt{3}$: $\lambda = \frac{(4 - \sqrt{6})(2\sqrt{2} + \sqrt{3})}{(2\sqrt{2} - \sqrt{3})(2\sqrt{2} + \sqrt{3})}$ $\lambda = \frac{8\sqrt{2} + 4\sqrt{3} - 2\sqrt{12} - \sqrt{18}}{8 - 3}$ $\lambda = \frac{8\sqrt{2} + 4\sqrt{3} - 4\sqrt{3} - 3\sqrt{2}}{5}$ $\lambda = \frac{5\sqrt{2}}{5}$ $\lambda = \sqrt{2}$

Step 8: Calculate $\alpha + 2\beta + \gamma$

We want to find $(\alpha + 2\beta + \gamma)^2$. We can rewrite $\alpha + 2\beta + \gamma$ as $(\alpha + \beta) + (\beta + \gamma)$. We know that $\alpha + \beta = 4\lambda$ and $\beta + \gamma = 3\sqrt{2}$. Since $\lambda = \sqrt{2}$, we have $\alpha + \beta = 4\sqrt{2}$. Therefore, $\alpha + 2\beta + \gamma = 4\sqrt{2} + 3\sqrt{2} = 7\sqrt{2}$.

Step 9: Calculate $(\alpha + 2\beta + \gamma)^2$

$(\alpha + 2\beta + \gamma)^2 = (7\sqrt{2})^2 = 49 \cdot 2 = 98$

Common Mistakes & Tips

• Careless algebraic manipulation is a common source of error. Double-check each step.
• Rationalizing the denominator is crucial for simplifying expressions with radicals.
• Look for ways to combine terms and simplify expressions before plugging in values for variables.

Summary

By applying Vieta's formulas, utilizing the given condition $\beta + \gamma = 3\sqrt{2}$, and simplifying the resulting expression, we found $\lambda = \sqrt{2}$. We then calculated $\alpha + 2\beta + \gamma = 7\sqrt{2}$ and $(\alpha + 2\beta + \gamma)^2 = 98$.

The final answer is \boxed{98}.