Let α and β be the roots of the equation x2+2ax+(3a+10)=0 such that α<1<β. Then the set of all possible values of a is :
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Official solution
Since α<1<β, the value of f(1) must be negative, where f(x)=x2+2ax+(3a+10):
\[f(1) < 0 \implies 1 + 2a + (3a+10) < 0 \implies 5a + 11 < 0 \implies a < -\frac{11}{5}\]
\[\therefore a \in \left(-\infty, -\frac{11}{5}\right)\]