Let $\alpha$ and $\beta$ be the roots of equation $x^{2} + 2ax + (3a + 10) = 0$ such that $\alpha &l

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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

$\because\alpha < 1 < \beta$$${f(1) < 0 }{\Rightarrow 1 + 2a + (3a + 10) < 0 }{\Rightarrow 5a + 11 < 0 }{a < \frac{- 11}{5} }{	herefore a \in \left( - \infty,\frac{- 11}{5} ight)}$$

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