Question

If \(\alpha, \beta\) are roots of the quadratic equation \[ \lambda x^2 - (\lambda+3)x + 3 = 0 \] and \(\alpha<\beta\) such that \[ \frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3}, \] then find the sum of all possible values of \(\lambda\).

• A. \(3\)
• B. \(2\)
• C. \(4\)
• D. \(6\)

Hint:

In quadratic root problems:

Convert expressions involving reciprocals into sums and products.
Use \((\beta-\alpha)^2=(\alpha+\beta)^2-4\alpha\beta\).
Absolute value equations usually give multiple parameter values.

Answer 1

The Correct Option is D

Concept: For a quadratic equation \(ax^2+bx+c=0\) with roots \(\alpha, \beta\): \[ \alpha+\beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a} \] Also, \[ \frac{1}{\alpha}-\frac{1}{\beta}=\frac{\beta-\alpha}{\alpha\beta} \] Step 1: Find sum and product of roots. Given: \[ \lambda x^2-(\lambda+3)x+3=0 \] \[ \alpha+\beta=\frac{\lambda+3}{\lambda}, \qquad \alpha\beta=\frac{3}{\lambda} \]
Step 2: Use the given condition. \[ \frac{1}{\alpha}-\frac{1}{\beta} =\frac{\beta-\alpha}{\alpha\beta} =\frac{1}{3} \] \[ \Rightarrow \frac{\beta-\alpha}{\frac{3}{\lambda}}=\frac{1}{3} \Rightarrow \beta-\alpha=\frac{1}{\lambda} \]
Step 3: Express \(\beta-\alpha\) using sum and product. \[ (\beta-\alpha)^2=(\alpha+\beta)^2-4\alpha\beta \] \[ (\beta-\alpha)^2 =\left(\frac{\lambda+3}{\lambda}\right)^2-\frac{12}{\lambda} =\frac{(\lambda-3)^2}{\lambda^2} \] \[ \Rightarrow \beta-\alpha=\frac{|\lambda-3|}{\lambda} \] From Step 2: \[ \frac{|\lambda-3|}{\lambda}=\frac{1}{\lambda} \Rightarrow |\lambda-3|=1 \]
Step 4: Find possible values of \(\lambda\). \[ \lambda-3=1 \Rightarrow \lambda=4 \] \[ \lambda-3=-1 \Rightarrow \lambda=2 \]
Step 5: Find the required sum. \[ \lambda_1+\lambda_2=4+2=6 \]
Conclusion: \[ \boxed{\text{Sum of all possible values of } \lambda = 6} \] Hence, the correct answer is (4).