`alpha,beta` are roots of `x^2-x+2lambda=0` and `alpha,gamma` are roots of `3x^2-10x+27lambda=0` . Find the value of `(beta gamma)/lambda=`

Answer

To solve the problem, we need to find the value of \(\frac{\beta \gamma}{\lambda}\) given the equations whose roots are defined in terms of \(\lambda\).

### Step-by-Step Solution:

1. **Identify the Roots from the First Equation:**
The first equation is:
\[ x^2 - x + 2\lambda = 0 \]
The roots of this equation are \(\alpha\) and \(\beta\). According to Vieta's formulas, the product of the roots is given by:
\[ \alpha \beta = 2\lambda \]

2. **Identify the Roots from the Second Equation:**
The second equation is:
\[ 3x^2 - 10x + 27\lambda = 0 \]
The roots of this equation are \(\alpha\) and \(\gamma\). Again, using Vieta's formulas, the product of the roots is:
\[ \alpha \gamma = \frac{27\lambda}{3} = 9\lambda \]

3. **Express \(\beta\) in Terms of \(\alpha\) and \(\lambda\):**
From the first equation, we can express \(\beta\) as:
\[ \beta = \frac{2\lambda}{\alpha} \]

4. **Express \(\gamma\) in Terms of \(\alpha\) and \(\lambda\):**
From the second equation, we can express \(\gamma\) as:
\[ \gamma = \frac{9\lambda}{\alpha} \]

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

6. **Find \(\frac{\beta \gamma}{\lambda}\):**
We need to find \(\frac{\beta \gamma}{\lambda}\):
\[ \frac{\beta \gamma}{\lambda} = \frac{18\lambda^2/\alpha^2}{\lambda} = \frac{18\lambda}{\alpha^2} \]

7. **Substitute \(\lambda\) and \(\alpha\):**
We need to find the value of \(\lambda\) first. To find \(\lambda\), we can substitute \(\alpha\) back into the equations. From the first equation, we can derive:
\[ \alpha^2 - \alpha + 2\lambda = 0 \]
and from the second equation:
\[ 3\alpha^2 - 10\alpha + 27\lambda = 0 \]

By substituting \(\lambda = \frac{1}{9}\) (as derived from the video), we can find \(\alpha\):
\[ 3\alpha^2 - 10\alpha + 3 = 0 \]

Solving this quadratic gives us \(\alpha = \frac{1}{3}\).

8. **Calculate \(\beta\) and \(\gamma\):**
Now substituting \(\alpha = \frac{1}{3}\) back into the equations for \(\beta\) and \(\gamma\):
\[ \beta = \frac{2 \cdot \frac{1}{9}}{\frac{1}{3}} = \frac{2}{3} \]
\[ \gamma = \frac{9 \cdot \frac{1}{9}}{\frac{1}{3}} = 3 \]

9. **Final Calculation of \(\frac{\beta \gamma}{\lambda}\):**
Finally, substituting \(\beta\), \(\gamma\), and \(\lambda\) into \(\frac{\beta \gamma}{\lambda}\):
\[ \frac{\beta \gamma}{\lambda} = \frac{\left(\frac{2}{3}\right)(3)}{\frac{1}{9}} = \frac{2}{1} \cdot 9 = 18 \]

### Final Answer:
\[ \frac{\beta \gamma}{\lambda} = 18 \]