JEE Main 2026 (21 January Evening Shift)

Question

Let $\alpha$ and $\beta$ be the roots of equation $x^2 + 2ax + (3a + 10) = 0$ such that $\alpha < 1 < \beta$. Then the set of all possible values of $a$ is:

Accepted Answer

$(-\infty, -\frac{11}{5})$

Answer

$(-\infty, -2) \cup (5, \infty)$

Answer

$(-\infty, -3)$

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