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Question

If $\alpha, \beta$ ($\alpha < \beta$) are the roots of the equation $\lambda x^2 - (\lambda+3)x + 3 = 0$ such that $\dfrac{1}{\alpha} - \dfrac{1}{\beta} = \dfrac13$, then the sum of all possible values of $\lambda$ is

Accepted Answer

Factor the equation: $$\lambda x^2 - \lambda x - 3x + 3 = 0 \implies \lambda x(x-1) - 3(x-1) = 0 \implies (x-1)(\lambda x - 3) = 0$$ so the roots are $x=1$ and $x = \dfrac3\lambda$. Case A: $\alpha=1,\ \beta=\dfrac3\lambda$ (needs $\beta>1$). $$\frac{1}{1} - \frac{\lambda}{3} = \frac13 \implies \frac{\lambda}{3} = \frac23 \implies \lambda = 2$$ Check: $\beta = 3/2 = 1.5 > 1$ --- valid. Case B: $\alpha=\dfrac3\lambda,\ \beta=1$ (needs $\alpha<1$). $$\frac{\lambda}{3} - 1 = \frac13 \implies \frac{\lambda}{3} = \frac43 \implies \lambda = 4$$ Check: $\alpha = 3/4 = 0.75 < 1$ --- valid. Sum of possible values of $\lambda$: $2 + 4 = 6$.

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