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:
  • Sum of roots: $r_1 + r_2 = -\frac{b}{a}$
  • Product of roots: $r_1 r_2 = \frac{c}{a}$
• Forming a Quadratic Equation: Given the sum (S) and product (P) of the roots, the quadratic equation is $x^2 - Sx + P = 0$.

Step-by-Step Solution

Step 1: Apply Vieta's Formulas to the Given Quadratic Equations

We apply Vieta's formulas to the two given quadratic equations to establish relationships between their roots and coefficients.

Equation 1: $14x^2 - 31x + 3\lambda = 0$ with roots $\alpha$ and $\beta$.

• Sum of roots: $\alpha + \beta = -\frac{-31}{14} = \frac{31}{14}$ (Equation 1.1)
• Product of roots: $\alpha \beta = \frac{3\lambda}{14}$ (Equation 1.2)

Equation 2: $35x^2 - 53x + 4\lambda = 0$ with roots $\alpha$ and $\gamma$.

• Sum of roots: $\alpha + \gamma = -\frac{-53}{35} = \frac{53}{35}$ (Equation 2.1)
• Product of roots: $\alpha \gamma = \frac{4\lambda}{35}$ (Equation 2.2)

Step 2: Establish a Relationship Between $\beta$ and $\gamma$

We aim to eliminate $\alpha$ and $\lambda$ from the product of roots equations to find a relationship between $\beta$ and $\gamma$. We divide Equation 1.2 by Equation 2.2:

$\frac{\alpha \beta}{\alpha \gamma} = \frac{\frac{3\lambda}{14}}{\frac{4\lambda}{35}}$

Since $\lambda \ne 0$ and $\alpha \ne 0$ (otherwise, $3\lambda = 0$ and $\lambda = 0$), we can cancel $\alpha$ and $\lambda$:

$\frac{\beta}{\gamma} = \frac{3\lambda}{14} \cdot \frac{35}{4\lambda} = \frac{3}{14} \cdot \frac{35}{4} = \frac{3 \cdot 5 \cdot 7}{2 \cdot 7 \cdot 4} = \frac{15}{8}$

Thus, $\beta = \frac{15}{8}\gamma$ (Equation 3)

Step 3: Solve for $\gamma$

We subtract Equation 2.1 from Equation 1.1 to eliminate $\alpha$:

$(\alpha + \beta) - (\alpha + \gamma) = \frac{31}{14} - \frac{53}{35}$ $\beta - \gamma = \frac{31}{14} - \frac{53}{35} = \frac{31 \cdot 5}{14 \cdot 5} - \frac{53 \cdot 2}{35 \cdot 2} = \frac{155}{70} - \frac{106}{70} = \frac{49}{70} = \frac{7}{10}$ (Equation 4)

Substitute Equation 3 into Equation 4:

$\frac{15}{8}\gamma - \gamma = \frac{7}{10}$ $\frac{7}{8}\gamma = \frac{7}{10}$ $\gamma = \frac{7}{10} \cdot \frac{8}{7} = \frac{8}{10} = \frac{4}{5}$

Step 4: Solve for $\beta$ and $\alpha$

Substitute $\gamma = \frac{4}{5}$ into Equation 3:

$\beta = \frac{15}{8} \cdot \frac{4}{5} = \frac{3 \cdot 5}{2 \cdot 4} \cdot \frac{4}{5} = \frac{3}{2}$

Substitute $\beta = \frac{3}{2}$ into Equation 1.1:

$\alpha + \frac{3}{2} = \frac{31}{14}$ $\alpha = \frac{31}{14} - \frac{3}{2} = \frac{31}{14} - \frac{21}{14} = \frac{10}{14} = \frac{5}{7}$

So, $\alpha = \frac{5}{7}$, $\beta = \frac{3}{2}$, and $\gamma = \frac{4}{5}$.

Step 5: Calculate the Value of $\lambda$ (Verification)

Using Equation 1.2:

$\alpha \beta = \frac{3\lambda}{14}$ $\frac{5}{7} \cdot \frac{3}{2} = \frac{3\lambda}{14}$ $\frac{15}{14} = \frac{3\lambda}{14}$ $\lambda = 5$

Step 6: Calculate the Sum of the New Roots

The new roots are $\frac{3\alpha}{\beta}$ and $\frac{4\alpha}{\gamma}$. Their sum, S, is:

$S = \frac{3\alpha}{\beta} + \frac{4\alpha}{\gamma} = \frac{3 \cdot \frac{5}{7}}{\frac{3}{2}} + \frac{4 \cdot \frac{5}{7}}{\frac{4}{5}} = \frac{\frac{15}{7}}{\frac{3}{2}} + \frac{\frac{20}{7}}{\frac{4}{5}} = \frac{15}{7} \cdot \frac{2}{3} + \frac{20}{7} \cdot \frac{5}{4} = \frac{5 \cdot 2}{7} + \frac{5 \cdot 5}{7} = \frac{10}{7} + \frac{25}{7} = \frac{35}{7} = 5$

Step 7: Calculate the Product of the New Roots

The product of the new roots, P, is:

$P = \frac{3\alpha}{\beta} \cdot \frac{4\alpha}{\gamma} = \frac{12\alpha^2}{\beta\gamma} = \frac{12 \cdot \left(\frac{5}{7}\right)^2}{\frac{3}{2} \cdot \frac{4}{5}} = \frac{12 \cdot \frac{25}{49}}{\frac{12}{10}} = \frac{12 \cdot \frac{25}{49}}{\frac{6}{5}} = 12 \cdot \frac{25}{49} \cdot \frac{5}{6} = \frac{12}{6} \cdot \frac{25 \cdot 5}{49} = 2 \cdot \frac{125}{49} = \frac{250}{49}$

Step 8: Form the Required Quadratic Equation

The quadratic equation is $x^2 - Sx + P = 0$, so:

$x^2 - 5x + \frac{250}{49} = 0$

Multiply by 49:

$49x^2 - 245x + 250 = 0$

Common Mistakes & Tips

• Double-check fraction arithmetic, especially when simplifying complex fractions.
• Remember the correct signs in Vieta's formulas ($-b/a$ for the sum).
• Verify that you are not dividing by zero when cancelling variables.

Summary

We used Vieta's formulas to relate the roots and coefficients of the given quadratic equations, solved for the roots $\alpha, \beta, \gamma$, calculated the sum and product of the new roots $\frac{3\alpha}{\beta}$ and $\frac{4\alpha}{\gamma}$, and constructed the quadratic equation $49x^2 - 245x + 250 = 0$.

The final answer is \boxed{49x^2 - 245x + 250 = 0}, which corresponds to option (B).