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
$$0>4x^2+4ax+4a^2+24a=(2x+a)^2+3a^2+24a$$
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.
We are interested in $f(x)=x^2+ax+a^2+6a<0\;\forall x \in (1,2)$
$$f(1)=1+a+a^2+6a \leq 0$$
That is $$a^2+7a+1\leq 0$$
$$ \frac{-7-\sqrt{45}}{2}\leq a \leq \frac{-7+\sqrt{45}}{2}$$
Also, we want, $$f(2)=4+2a+a^2+6a \leq 0$$
$$a^2+8a+4 \leq 0$$
$$\frac{-8-\sqrt{48}}{2} \leq a \leq \frac{-8+\sqrt{48}}{2}$$
Hence, overall, we need to take the intersection of the two constraint.$$ \frac{-7-\sqrt{45}}{2}\leq a \leq \frac{-8+\sqrt{48}}{2}$$
Rewrite the inequality as $$(2a+x+6)^2< (6-x)(3x+6)$$ The right hand side is positive for $1<x<2$, so we get $$-\sqrt{(6-x)(3x+6)}<2a+x+6<\sqrt{(6-x)(3x+6)}$$ and $$-x-6-\sqrt{(6-x)(3x+6)}<2a<-x-6+\sqrt{(6-x)(3x+6)}$$ $$\frac{-x-6-\sqrt{(6-x)(3x+6)}}{2}<a<\frac{-x-6+\sqrt{(6-x)(3x+6)}}{2}$$ The maximum of the right hand side is $-4+2 \sqrt{3}$ (let $x=2$) and the minimum of the left side is $-\frac{7}{2}-\frac{3 \sqrt{5}}{2}$ (let $x=1$) so $$-\frac{7}{2}-\frac{3 \sqrt{5}}{2}<a<-4+2 \sqrt{3}$$
You are on the right track.
$$\left(x+\frac{a}{2}\right)^2+\frac{3a^2+24a}{4}<0$$
Now observe that $\left(x+\frac{a}{2}\right)^2$ is always non-negative. Hence the above inequality reduces to: $$\frac{3a^2+24a}{4}<0$$ $$\implies 3a^2+24a<0$$ $$\implies 3a(a+8)<0$$
Now a product of two factors can be negative iff one of them is positive and one is negative.
So either $a<0$ and $a>-8$ $\implies -8<a<0$
or $a>0$ and $a<-8$ $\implies a\in (-\infty,-8)\cap(0,\infty)$
Now check for $a=1$. You will see that the above inequality does not hold whereas it does if $a=-1$.
Hence the required values of $a$ belong to the interval $\color{red}{-8<a<0}$.
Hint:-
If $x$ is either $1$ or $2$ as mentioned in your question,we have the whole squared part $(x+\frac{a}{2})$ is always $>0$.So, for the whole thing to be less than $0$, the $\frac{3a^2+24a}{4}$ part must be $<0$
