Quadratic EquationHard
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 :
Options
A.$\left( - \infty,\frac{- 11}{5} \right) \cup (5,\infty)$
B.$( - \infty, - 2) \cup (5,\infty)$
C.$( - \infty, - 3)$
D.$\left( - \infty,\frac{- 11}{5} \right)$
Solution
$\because\alpha < 1 < \beta$
$${f(1) < 0 }{\Rightarrow 1 + 2a + (3a + 10) < 0 }{\Rightarrow 5a + 11 < 0 }{a < \frac{- 11}{5} }{\therefore a \in \left( - \infty,\frac{- 11}{5} \right)}$$
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