Question

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

• A. \(8\)
• B. \(6\)
• C. \(4\)
• D. \(2\)

Hint:

In root-based problems, converting reciprocal conditions into expressions involving sum and product simplifies the algebra.

Answer 1

The Correct Option is B

Concept: Use relations between roots and coefficients and manipulate the given condition involving reciprocals.
Step 1: Use Vieta’s formulas \[ \alpha+\beta=\frac{\lambda+3}{\lambda},\quad \alpha\beta=\frac{3}{\lambda} \]
Step 2: Use the given condition \[ \frac{1}{\alpha}-\frac{1}{\beta}=\frac{\beta-\alpha}{\alpha\beta}=\frac{1}{3} \] \[ \Rightarrow \beta-\alpha=\frac{\alpha\beta}{3}=\frac{1}{\lambda} \]
Step 3: Square both sides \[ (\beta-\alpha)^2=(\alpha+\beta)^2-4\alpha\beta \] \[ \frac{1}{\lambda^2}=\left(\frac{\lambda+3}{\lambda}\right)^2-\frac{12}{\lambda} \] \[ \Rightarrow (\lambda-3)^2=13 \]
Step 4: Find possible values of \( \lambda \) \[ \lambda=3\pm\sqrt{13} \]
Step 5: Find the required sum \[ (3+\sqrt{13})+(3-\sqrt{13})=6 \]