If `alpha,beta` are roots of `x^2-x+2lambda=0` and `alpha,gamma` are roots of `3x^2-10x+27lambda=0` then value of `(betagamma)/(lambda)` is

Options

A. 27
B. 18
C. 9
D. 15

Answer

To solve the problem, we need to find the value of \(\frac{\beta \gamma}{\lambda}\) given the equations:

1. \(x^2 - x + 2\lambda = 0\) (roots are \(\alpha\) and \(\beta\))
2. \(3x^2 - 10x + 27\lambda = 0\) (roots are \(\alpha\) and \(\gamma\))

### Step 1: Find the product of the roots \(\alpha\) and \(\beta\)

From the first equation \(x^2 - x + 2\lambda = 0\), we can use Vieta's formulas which state that the product of the roots of a quadratic equation \(ax^2 + bx + c = 0\) is given by \(\frac{c}{a}\).

Thus, for the first equation:
\[ \alpha \beta = 2\lambda \]

### Step 2: Find the product of the roots \(\alpha\) and \(\gamma\)

From the second equation \(3x^2 - 10x + 27\lambda = 0\), again using Vieta's formulas:
\[ \alpha \gamma = \frac{27\lambda}{3} = 9\lambda \]

### Step 3: Express \(\beta\) in terms of \(\alpha\) and \(\lambda\)

From \(\alpha \beta = 2\lambda\), we can express \(\beta\) as:
\[ \beta = \frac{2\lambda}{\alpha} \]

### Step 4: Express \(\gamma\) in terms of \(\alpha\) and \(\lambda\)

From \(\alpha \gamma = 9\lambda\), we can express \(\gamma\) as:
\[ \gamma = \frac{9\lambda}{\alpha} \]

### Step 5: Calculate \(\beta \gamma\)

Now we can find \(\beta \gamma\):
\[ \beta \gamma = \left(\frac{2\lambda}{\alpha}\right) \left(\frac{9\lambda}{\alpha}\right) = \frac{18\lambda^2}{\alpha^2} \]

### Step 6: Find \(\frac{\beta \gamma}{\lambda}\)

Now, we need to compute \(\frac{\beta \gamma}{\lambda}\):
\[ \frac{\beta \gamma}{\lambda} = \frac{18\lambda^2/\alpha^2}{\lambda} = \frac{18\lambda}{\alpha^2} \]

### Step 7: Find the value of \(\lambda\)

To find \(\lambda\), we need to substitute \(\alpha\) back into the equations. We can find \(\alpha\) from the first equation:
\[ \alpha^2 - \alpha + 2\lambda = 0 \]
Substituting \(\lambda = \frac{1}{9}\) into the equation gives:
\[ \alpha^2 - \alpha + \frac{2}{9} = 0 \]

Using the quadratic formula:
\[ \alpha = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot \frac{2}{9}}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 - \frac{8}{9}}}{2} = \frac{1 \pm \sqrt{\frac{1}{9}}}{2} = \frac{1 \pm \frac{1}{3}}{2} \]
This gives:
\[ \alpha = \frac{2}{3} \quad \text{or} \quad \alpha = \frac{1}{3} \]

### Step 8: Calculate \(\frac{\beta \gamma}{\lambda}\)

Substituting \(\lambda = \frac{1}{9}\) and \(\alpha = \frac{1}{3}\) into \(\frac{\beta \gamma}{\lambda}\):
\[ \frac{\beta \gamma}{\lambda} = \frac{18 \cdot \frac{1}{9}}{\left(\frac{1}{3}\right)^2} = \frac{2}{\frac{1}{9}} = 18 \]

Thus, the final answer is:
\[ \frac{\beta \gamma}{\lambda} = 18 \]