Question

Let \(\lambda \neq 0\) be in R. If \(\alpha\) and \(\beta\) are the roots of the equation \(x^2 - x + 2\lambda = 0\), and \(\alpha\) and \(\gamma\) are the roots of the equation \(3x^2 - 10x + 27\lambda = 0\), then \(\frac{\beta\gamma}{\lambda}\) is equal to ___________.

Hint:

When two quadratic equations have a common root, a standard procedure is to eliminate the \(x^2\) term to get a linear relation between the root and the parameters. Substituting this back into one of the original equations then solves for the parameters.

Answer 1

Correct Answer: 18

Step 1: Understanding the Question
We are given two quadratic equations with a common root \(\alpha\). We need to find the value of an expression involving the other roots \(\beta, \gamma\) and the parameter \(\lambda\).
Step 2: Key Formula or Approach
We will use Vieta's formulas for the sum and product of roots. Then, we will use the fact that the common root \(\alpha\) must satisfy both equations to find the value of \(\lambda\) and \(\alpha\).
Step 3: Detailed Explanation
From the first equation, \(x^2 - x + 2\lambda = 0\): Sum of roots: \(\alpha + \beta = 1\) (1) Product of roots: \(\alpha\beta = 2\lambda\) (2) From the second equation, \(3x^2 - 10x + 27\lambda = 0\): Sum of roots: \(\alpha + \gamma = \frac{10}{3}\) (3) Product of roots: \(\alpha\gamma = \frac{27\lambda}{3} = 9\lambda\) (4) Since \(\alpha\) is a root of both equations, it must satisfy them:
\(\alpha^2 - \alpha + 2\lambda = 0\) (5)
\(3\alpha^2 - 10\alpha + 27\lambda = 0\) (6)
Multiply equation (5) by 3:
\(3\alpha^2 - 3\alpha + 6\lambda = 0\) (7)
Subtract equation (7) from equation (6):
\((3\alpha^2 - 10\alpha + 27\lambda) - (3\alpha^2 - 3\alpha + 6\lambda) = 0\)
\(-7\alpha + 21\lambda = 0 \implies 7\alpha = 21\lambda \implies \alpha = 3\lambda\).
Now substitute \(\alpha = 3\lambda\) into equation (5):
\((3\lambda)^2 - (3\lambda) + 2\lambda = 0\)
\(9\lambda^2 - \lambda = 0 \implies \lambda(9\lambda - 1) = 0\).
Since we are given \(\lambda \neq 0\), we must have \(9\lambda - 1 = 0 \implies \lambda = \frac{1}{9}\).
With \(\lambda = 1/9\), we find \(\alpha = 3\lambda = 3(1/9) = 1/3\).
Now we can find \(\beta\) and \(\gamma\).
From (1): \(1/3 + \beta = 1 \implies \beta = 2/3\).
From (3): \(1/3 + \gamma = 10/3 \implies \gamma = 9/3 = 3\).
Finally, we calculate the required expression:
\[ \frac{\beta\gamma}{\lambda} = \frac{(2/3)(3)}{1/9} = \frac{2}{1/9} = 18 \] Step 4: Final Answer
The value of the expression is 18.