Key Concepts and Formulas

• Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, the sum of the roots is $\alpha + \beta = -\frac{b}{a}$ and the product of the roots is $\alpha \beta = \frac{c}{a}$.
• Common Root: If two quadratic equations share a common root, that root must satisfy both equations.
• Solving Simultaneous Equations: Methods for solving simultaneous equations can be used to find unknown variables.

Step-by-Step Solution

Step 1: Applying 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. This allows us to establish relationships between $\alpha$, $\beta$, and $\lambda$.

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

Step 2: Applying 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, establishing relationships between $\alpha$, $\gamma$, and $\lambda$.

• 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)

Step 3: Finding the Common Root ($\alpha$)

Since $\alpha$ is a root of both equations, it must satisfy both:

$$\alpha^2 - \alpha + 2\lambda = 0$$ $$3\alpha^2 - 10\alpha + 27\lambda = 0$$

Multiplying the first equation by $27/2$ gives

$$\frac{27}{2}\alpha^2 - \frac{27}{2}\alpha + 27\lambda = 0$$

Subtracting the second equation from this gives

$$\frac{27}{2}\alpha^2 - \frac{27}{2}\alpha - (3\alpha^2 - 10\alpha) = 0$$ $$(\frac{27}{2}-3)\alpha^2 + (10-\frac{27}{2})\alpha = 0$$ $$\frac{21}{2}\alpha^2 - \frac{7}{2}\alpha = 0$$ $$\alpha(3\alpha - 1) = 0$$

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

If $\alpha = 0$, then from Equation 2, $\alpha \beta = 2\lambda$, so $0 = 2\lambda$, which implies $\lambda = 0$. However, we are given that $\lambda \ne 0$, so $\alpha$ cannot be 0.

Therefore, the common root is $\alpha = \frac{1}{3}$.

Step 4: Finding the Value of $\lambda$

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

$$\alpha \beta = 2\lambda$$

We don't know $\beta$ yet, so we can't directly use this. Instead, substitute $\alpha = \frac{1}{3}$ into the first quadratic equation:

$$(\frac{1}{3})^2 - \frac{1}{3} + 2\lambda = 0$$ $$\frac{1}{9} - \frac{1}{3} + 2\lambda = 0$$ $$\frac{1}{9} - \frac{3}{9} + 2\lambda = 0$$ $$-\frac{2}{9} + 2\lambda = 0$$ $$2\lambda = \frac{2}{9}$$ $$\lambda = \frac{1}{9}$$

So, $\lambda = \frac{1}{9}$.

Step 5: Finding the Roots $\beta$ and $\gamma$

Using Equation 1 ($\alpha + \beta = 1$) and the value $\alpha = \frac{1}{3}$:

$$\frac{1}{3} + \beta = 1$$ $$\beta = 1 - \frac{1}{3} = \frac{2}{3}$$

Using Equation 3 ($\alpha + \gamma = \frac{10}{3}$) and the value $\alpha = \frac{1}{3}$:

$$\frac{1}{3} + \gamma = \frac{10}{3}$$ $$\gamma = \frac{10}{3} - \frac{1}{3} = \frac{9}{3} = 3$$

So, we have found $\beta = \frac{2}{3}$ and $\gamma = 3$.

Step 6: Calculating the Target Expression $\frac{\beta \gamma}{\lambda}$

Substitute the values of $\beta$, $\gamma$, and $\lambda$ into the expression:

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

Common Mistakes & Tips

• Don't assume $\alpha = 0$: Always check if discarding $\alpha = 0$ is valid based on the problem's constraints.
• Careful with signs: Pay close attention to signs when applying Vieta's formulas, especially with negative coefficients.

Summary

This problem involved using Vieta's formulas to relate the roots and coefficients of two quadratic equations that share a common root. By solving the system of equations, we found the value of the common root, $\alpha$, and the parameter $\lambda$. We then determined the other roots, $\beta$ and $\gamma$, and calculated the desired expression $\frac{\beta \gamma}{\lambda}$, which equals 18.

The final answer is \boxed{18}.