Let $\alpha,\beta$ be the roots of the quadratic equation $12x^{2} - 20x + 3\lambda = 0,\lambda \in

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Let $\alpha,\beta$ be the roots of the quadratic equation $12x^{2} - 20x + 3\lambda = 0,\lambda \in \mathbb{Z}$. If $\frac{1}{2} \leq |\beta - \alpha| \leq \frac{3}{2}$, then the sum of all possible values of $\lambda$ is :

Accepted Answer

$\ \frac{1}{2} \leq |\alpha - \beta| \leq \frac{3}{2}$

$${\frac{1}{4} \leq |\alpha - \beta|^{2} \leq \frac{9}{4} }{\frac{1}{4} \leq (\alpha + \beta)^{2} - 4\alpha\beta \leq \frac{9}{4} }{\frac{1}{4} \leq \frac{25}{9} - 4 \times \frac{\lambda}{4} \leq \frac{9}{4} }{- \frac{91}{36} \leq - \lambda \leq \frac{- 19}{36} }{\frac{19}{36} \leq \lambda \leq \frac{91}{36} }{\lambda = 1,2 }$$Sum $= 3$

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