Let `lambda != 0` be a real number.Let `alpha,beta` be the roots of the equation `14x^(2)-31x+3 lambda=0` and `alpha,gamma` be the roots of the equation `35x^(2)-53x+4 lambda=0`.Then `(3 alpha)/(beta)` and `(4 alpha)/(gamma)` are the roots of the equation

Options

A. `7x^(2)+245x-250=0`
B. `49x^(2)+245x+250=0`
C. `49x^(2)-245x+250=0`
D. `7x^(2)-245x+250=0`

Answer

To solve the problem, we need to find the roots of the equations given and then derive a new equation based on the specified transformations of the roots. Let's break it down step by step.

### Step 1: Identify the roots of the first equation
The first equation is given by:
\[ 14x^2 - 31x + 3\lambda = 0 \]

Using Vieta's formulas, we can find:
- The sum of the roots (\(\alpha + \beta\)):
\[ \alpha + \beta = \frac{31}{14} \]
- The product of the roots (\(\alpha \beta\)):
\[ \alpha \beta = \frac{3\lambda}{14} \]

### Step 2: Identify the roots of the second equation
The second equation is given by:
\[ 35x^2 - 53x + 4\lambda = 0 \]

Again, using Vieta's formulas:
- The sum of the roots (\(\alpha + \gamma\)):
\[ \alpha + \gamma = \frac{53}{35} \]
- The product of the roots (\(\alpha \gamma\)):
\[ \alpha \gamma = \frac{4\lambda}{35} \]

### Step 3: Find the ratio of \(\beta\) and \(\gamma\)
To find the ratio \(\frac{\beta}{\gamma}\), we can use the products of the roots from the two equations:
\[ \frac{\alpha \beta}{\alpha \gamma} = \frac{\frac{3\lambda}{14}}{\frac{4\lambda}{35}} = \frac{3 \cdot 35}{4 \cdot 14} = \frac{105}{56} = \frac{15}{8} \]
Thus, we have:
\[ \frac{\beta}{\gamma} = \frac{15}{8} \]

### Step 4: Find \(\beta - \gamma\)
Now, we can find \(\beta - \gamma\) using the sums of the roots:
\[ \beta - \gamma = (\alpha + \beta) - (\alpha + \gamma) = \frac{31}{14} - \frac{53}{35} \]
Finding a common denominator (which is 70):
\[ \beta - \gamma = \frac{31 \cdot 5}{70} - \frac{53 \cdot 2}{70} = \frac{155 - 106}{70} = \frac{49}{70} = \frac{7}{10} \]

### Step 5: Solve for \(\beta\) and \(\gamma\)
Let’s denote \(\beta = 15k\) and \(\gamma = 8k\) for some \(k\). From \(\beta - \gamma = \frac{7}{10}\):
\[ 15k - 8k = \frac{7}{10} \implies 7k = \frac{7}{10} \implies k = \frac{1}{10} \]
Thus:
\[ \beta = 15 \cdot \frac{1}{10} = \frac{3}{2}, \quad \gamma = 8 \cdot \frac{1}{10} = \frac{4}{5} \]

### Step 6: Find \(\alpha\)
Using \(\alpha + \beta = \frac{31}{14}\):
\[ \alpha + \frac{3}{2} = \frac{31}{14} \implies \alpha = \frac{31}{14} - \frac{21}{14} = \frac{10}{14} = \frac{5}{7} \]

### Step 7: Form the new roots
Now we need to find the new roots:
\[ \frac{3\alpha}{\beta} = \frac{3 \cdot \frac{5}{7}}{\frac{3}{2}} = \frac{30}{21} = \frac{10}{7} \]
\[ \frac{4\alpha}{\gamma} = \frac{4 \cdot \frac{5}{7}}{\frac{4}{5}} = \frac{20}{7} \cdot \frac{5}{4} = \frac{25}{7} \]

### Step 8: Form the equation with these roots
The roots are \(\frac{10}{7}\) and \(\frac{25}{7}\). The sum and product of the roots are:
- Sum:
\[ \frac{10}{7} + \frac{25}{7} = \frac{35}{7} = 5 \]
- Product:
\[ \frac{10}{7} \cdot \frac{25}{7} = \frac{250}{49} \]

Thus, the equation with roots \(\frac{10}{7}\) and \(\frac{25}{7}\) is:
\[ x^2 - 5x + \frac{250}{49} = 0 \]
Multiplying through by 49 to eliminate the fraction:
\[ 49x^2 - 245x + 250 = 0 \]

### Final Answer
The final equation is:
\[ 49x^2 - 245x + 250 = 0 \]