<s>[INST] Quadratic Inequality in terms of variable $x$ 
Find the values of $a$ for which the inequality $x^2+ax+a^2+6a<0\;\forall x \in (1,2)$

$\bf{My\; Try::}$ We can Write Equation as $$x^2+ax+\frac{a^2}{4}+\frac{3a^2}{4}+6a<0$$
So $$\left(x+\frac{a}{2}\right)^2+\frac{3a^2+24a}{4}<0$$
Now how can i solve after that, Help required, Thanks
 [/INST] $$0>4x^2+4ax+4a^2+24a=(2x+a)^2+3a^2+24a$$
Now $1<x<2\iff2+a<2x+a<4+a$
$\implies(2x+a)^2<$min$\{(2+a)^2,(4+a)^2\}$
Case$\#1:$ If $(4+a)^2\ge(2+a)^2\iff a\ge-3,(2x+a)^2<(2+a)^2$
$$0>(2+a)^2+3a^2+24a=4(a^2+7a+1)$$
We need $a^2+7a+1<0$
As the roots of $a^2+7a+1=0$ are $\dfrac{-7\pm\sqrt{49-4}}2=\dfrac{-7\pm3\sqrt5}2$
$\implies$ either $a>\dfrac{-7+3\sqrt5}2$ or $a<\dfrac{-7-3\sqrt5}2$
But $a\ge-3$ and $\dfrac{-7-3\sqrt5}2<-3$
Case$\#2:$ What if $(4+a)^2<(2+a)^2\iff a<-3$
Left for you.
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