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 $r_1r_2 = \frac{c}{a}$.
• Solving System of Equations: Techniques to solve for unknowns in multiple equations, such as substitution or elimination.

Step-by-Step Solution

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

We are given the equation $x^2 - x + 2\lambda = 0$ with roots $\alpha$ and $\beta$. We apply Vieta's formulas to relate the roots to the coefficients.

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

Reasoning: We use Vieta's formulas to establish a relationship between the roots and coefficients of the equation, which will be useful later in forming a system of equations.

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

We are given the equation $3x^2 - 10x + 27\lambda = 0$ with roots $\alpha$ and $\gamma$. We apply Vieta's formulas to relate the roots to the coefficients.

• Sum of roots: $\alpha + \gamma = -\frac{-10}{3} = \frac{10}{3}$ (Equation 3)
• Product of roots: $\alpha \gamma = \frac{27\lambda}{3} = 9\lambda$ (Equation 4)

Reasoning: Similar to Step 1, we are applying Vieta's formulas to the second equation to establish another set of relationships between its roots and coefficients.

Step 3: Express $\beta$ and $\gamma$ in terms of $\alpha$

From Equations 1 and 3, we can express $\beta$ and $\gamma$ in terms of $\alpha$.

• From Equation 1: $\beta = 1 - \alpha$
• From Equation 3: $\gamma = \frac{10}{3} - \alpha$

Reasoning: Expressing $\beta$ and $\gamma$ in terms of $\alpha$ allows us to substitute these expressions into Equations 2 and 4, ultimately leading to a system of equations with two unknowns, $\alpha$ and $\lambda$.

Step 4: Substitute $\beta$ and $\gamma$ into Equations 2 and 4

Substitute $\beta = 1 - \alpha$ into Equation 2, $\alpha\beta = 2\lambda$: $\alpha(1 - \alpha) = 2\lambda \implies \alpha - \alpha^2 = 2\lambda$ (Equation 5)

Substitute $\gamma = \frac{10}{3} - \alpha$ into Equation 4, $\alpha\gamma = 9\lambda$: $\alpha(\frac{10}{3} - \alpha) = 9\lambda \implies \frac{10}{3}\alpha - \alpha^2 = 9\lambda$ (Equation 6)

Reasoning: This substitution gives us two equations (Equations 5 and 6) containing only $\alpha$ and $\lambda$. We can then solve for these unknowns.

Step 5: Solve for $\alpha$ by Eliminating $\lambda$

Multiply Equation 5 by 9 and Equation 6 by 2 to eliminate $\lambda$:

• $9(\alpha - \alpha^2) = 18\lambda \implies 9\alpha - 9\alpha^2 = 18\lambda$
• $2(\frac{10}{3}\alpha - \alpha^2) = 18\lambda \implies \frac{20}{3}\alpha - 2\alpha^2 = 18\lambda$

Equate the left-hand sides: $9\alpha - 9\alpha^2 = \frac{20}{3}\alpha - 2\alpha^2$ $9\alpha - \frac{20}{3}\alpha = 9\alpha^2 - 2\alpha^2$ $\frac{7}{3}\alpha = 7\alpha^2$ $7\alpha^2 - \frac{7}{3}\alpha = 0$ $7\alpha(\alpha - \frac{1}{3}) = 0$

Therefore, $\alpha = 0$ or $\alpha = \frac{1}{3}$.

Reasoning: Eliminating $\lambda$ allows us to solve for $\alpha$. Factoring the resulting quadratic equation yields two possible solutions for $\alpha$.

Step 6: Determine the Correct Value of $\alpha$

Since $\lambda \ne 0$, $\alpha$ cannot be 0. If $\alpha = 0$, then from $\alpha \beta = 2\lambda$, we have $0 = 2\lambda$, implying $\lambda = 0$, which contradicts the given condition. Thus, $\alpha = \frac{1}{3}$.

Reasoning: The problem statement specifies that $\lambda$ cannot be zero. We use this condition to eliminate one of the solutions for $\alpha$.

Step 7: Solve for $\lambda$

Substitute $\alpha = \frac{1}{3}$ into Equation 5: $\frac{1}{3} - (\frac{1}{3})^2 = 2\lambda$ $\frac{1}{3} - \frac{1}{9} = 2\lambda$ $\frac{2}{9} = 2\lambda$ $\lambda = \frac{1}{9}$

Reasoning: Now that we have found $\alpha$, we substitute it back into one of the equations relating $\alpha$ and $\lambda$ to solve for $\lambda$.

Step 8: Solve for $\beta$ and $\gamma$

Substitute $\alpha = \frac{1}{3}$ into $\beta = 1 - \alpha$: $\beta = 1 - \frac{1}{3} = \frac{2}{3}$

Substitute $\alpha = \frac{1}{3}$ into $\gamma = \frac{10}{3} - \alpha$: $\gamma = \frac{10}{3} - \frac{1}{3} = \frac{9}{3} = 3$

Reasoning: We use the value of $\alpha$ to find the values of $\beta$ and $\gamma$.

Step 9: Calculate $\frac{\beta\gamma}{\lambda}$

Substitute $\beta = \frac{2}{3}$, $\gamma = 3$, and $\lambda = \frac{1}{9}$ into the expression:

$\frac{\beta\gamma}{\lambda} = \frac{(\frac{2}{3})(3)}{\frac{1}{9}} = \frac{2}{\frac{1}{9}} = 2 \times 9 = 18$

Reasoning: Finally, we substitute all the values we found into the expression we want to evaluate.

Common Mistakes & Tips

• Sign Errors: Double-check the signs when applying Vieta's formulas, particularly when $b$ or $c$ are negative.
• Algebraic Errors: Be careful with fractions and algebraic manipulations, especially when solving systems of equations.
• Condition on $\lambda$: Remember to check if your solution satisfies all given conditions, such as $\lambda \neq 0$.

Summary

We used Vieta's formulas to establish relationships between the roots and coefficients of the given quadratic equations. We then formed a system of equations, solved for the common root $\alpha$ and $\lambda$, and subsequently found the values of $\beta$ and $\gamma$. Finally, we calculated the value of the expression $\frac{\beta\gamma}{\lambda}$.

The final answer is \boxed{18}, which corresponds to option (D).