Let , be the roots of the quadratic equation 12x^2 - 20x + 3 = 0, Z.

Question

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

Accepted Answer

From the equation, $\alpha+\beta = \dfrac{20}{12} = \dfrac{5}{3}$ and $\alpha\beta = \dfrac{3\lambda}{12} = \dfrac{\lambda}{4}$.

$$\frac{1}{2} \le |\alpha-\beta| \le \frac{3}{2} \implies \frac{1}{4} \le (\alpha-\beta)^2 \le \frac{9}{4} \implies \frac{1}{4} \le (\alpha+\beta)^2 - 4\alpha\beta \le \frac{9}{4}$$

$$\frac{1}{4} \le \frac{25}{9} - \lambda \le \frac{9}{4}$$

Solving, $\dfrac{19}{36} \le \lambda \le \dfrac{91}{36}$, i.e. $\lambda \approx 0.53$ to $2.53$.

Integer values: $\lambda = 1, 2$.

Sum $= 1+2 = 3$.

Solutions · 1