Question

Given \(\frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3}\) such that roots of the quadratic equation \(\lambda x^2 + (\lambda+1)x + 3 = 0\) are \(\alpha\) & \(\beta\), then sum of values of \(\lambda\) is equal to :

• A. 8
• B. 9
• C. 10
• D. 11

Hint:

Whenever you encounter the difference of roots or reciprocals, always look for symmetric expressions like \(\alpha+\beta\) and \(\alpha\beta\). Squaring is usually the fastest way to bridge the gap.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The problem relates the coefficients of a quadratic equation to its roots using Vieta's formulas and a given algebraic condition involving the reciprocals of the roots.
Step 2: Key Formula or Approach:
For a quadratic equation \(ax^2 + bx + c = 0\) with roots \(\alpha\) and \(\beta\):
1. \(\alpha + \beta = -\frac{b}{a}\)
2. \(\alpha \beta = \frac{c}{a}\)
3. Identity: \((\beta - \alpha)^2 = (\alpha + \beta)^2 - 4\alpha\beta\)
Step 3: Detailed Explanation:
Given the equation \(\lambda x^2 + (\lambda+1)x + 3 = 0\), the sum and product of roots are:
\[ \alpha + \beta = -\frac{\lambda+1}{\lambda} \text{ and } \alpha \beta = \frac{3}{\lambda} \] The given condition is:
\[ \frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3} \implies \frac{\beta - \alpha}{\alpha \beta} = \frac{1}{3} \implies \beta - \alpha = \frac{\alpha \beta}{3} \] Substitute \(\alpha \beta = \frac{3}{\lambda}\):
\[ \beta - \alpha = \frac{3/\lambda}{3} = \frac{1}{\lambda} \] Squaring both sides to eliminate the difference of roots:
\[ (\beta - \alpha)^2 = \frac{1}{\lambda^2} \implies (\alpha + \beta)^2 - 4\alpha\beta = \frac{1}{\lambda^2} \] Substituting the expressions in terms of \(\lambda\):
\[ \left( -\frac{\lambda+1}{\lambda} \right)^2 - 4\left( \frac{3}{\lambda} \right) = \frac{1}{\lambda^2} \] \[ \frac{\lambda^2 + 2\lambda + 1}{\lambda^2} - \frac{12}{\lambda} = \frac{1}{\lambda^2} \] Multiplying the entire equation by \(\lambda^2\) (where \(\lambda \neq 0\)):
\[ \lambda^2 + 2\lambda + 1 - 12\lambda = 1 \] \[ \lambda^2 - 10\lambda = 0 \implies \lambda(\lambda - 10) = 0 \] Since it is a quadratic equation, \(\lambda\) cannot be zero (\(\lambda \neq 0\)). Thus, the only valid value is \(\lambda = 10\).
Step 4: Final Answer:
The sum of values of \(\lambda\) (since only one value exists) is 10.