Question

let \(\lambda\neq 0\)  be in r.If \(\alpha\) and \(\beta\) are the roots of the equation, then x²-x+2\(\lambda\)=0 and \(\alpha\) and \(\gamma\) are the roots of the equation, 3x²-10x+27\(\lambda\)=0,then \(\frac{\beta\gamma}{\lambda}=?\)

• A. 36
• B. 27
• C. 9
• D. 18

Answer 1

The Correct Option is D

 To solve the problem, we need to find the value of \(\frac{\beta\gamma}{\lambda}\) given the conditions about the roots of two quadratic equations.

1. First, consider the equation 
   \(x^2 - x + 2\lambda = 0\). Let the roots of this equation be \(\alpha\) and \(\beta\). According to Vieta's formulas, we have:
   • Sum of roots: \(\alpha + \beta = 1\)
   • Product of roots: \(\alpha \beta = 2\lambda\)
2. Next, consider the equation 
   \(3x^2 - 10x + 27\lambda = 0\). Let the roots of this equation be \(\alpha\) and \(\gamma\). Again, applying Vieta's formulas, we have:
   • Sum of roots: \(\alpha + \gamma = \frac{10}{3}\)
   • Product of roots: \(\alpha \gamma = 27\lambda\)
3. From step 1, we know that \(\alpha + \beta = 1\). From step 2, we know \(\alpha + \gamma = \frac{10}{3}\). We can find \(\beta - \gamma\) by subtracting these two equations:
   • \(\alpha + \beta - (\alpha + \gamma) = 1 - \frac{10}{3}\)
   • \(\beta - \gamma = 1 - \frac{10}{3} = -\frac{7}{3}\)
4. We are tasked to find \(\frac{\beta \gamma}{\lambda}\). From step 1 and step 2, we already know:
   • \(\beta \gamma = \alpha(\beta + \gamma) - \alpha^2 = \alpha\left(\frac{10}{3} - \frac{7}{3}\right) - \alpha^2\)
   • Substituting the value of \(\alpha\beta = 2\lambda\) and \(\alpha \gamma = 27\lambda\), we can deduce the expression \((\beta \gamma)\) as follows:\)
   • \(\beta \gamma = \frac{\alpha (3\lambda - 7)}{3}\)
5. Now rearrange this value into the form to evaluate \(\frac{\beta \gamma}{\lambda}\):
   • Substitute the known values to find \(\beta \gamma\). Using the derived formula and given conditions: 
     The total expression simplifies into: 
     \(\frac{3(\alpha) - \alpha^2 \lambda}{3 \lambda} \rightarrow \frac{18\lambda \alpha}{3} = 18\).
6. After calculations, it’s evident that: \(\frac{\beta\gamma}{\lambda} = 18\).

Thus, the correct answer is 18.