Find values of $a$ such that $x^2+ax+a^2+6a \lt 0$ $\forall$ $x \in (1,2)$

Question

Find values of $a$ such that $x^2+ax+a^2+6a \lt 0$ $\forall$ $x \in (1,2)$

My try: Since $y=x^2+ax+a^2+6a$ is an open upward Parabola, the roots $\alpha,\beta$ should be distinct and satisfy $1 \lt \alpha \lt 2$ and $1 \lt \beta \lt 2$ implies both the roots are distinct and lies between $1$ and $2$.

So we have the following conditions:

$1.$ Since it should cut positive Y axis as both the roots are positive we have $a^2+6a \gt 0$

$2.$ $D \gt 0$ $\implies$ $a^2+8a \lt 0$

$3.$ $f(1) \gt 0$ $\implies$ $a^2+7a+1 \gt 0$

$4.$ $f(2) \gt 0$ $\implies$ $a^2+8a+4 \gt 0$

How to solve all these four inequalities in a simpler way?

Answer 1

Hint: if you reverse the role of $a$ and $x$ you get $$a^2+a(x+6)+x^2<0$$

Solution of coresponding equation (on $a$) is $$a_{1,2} = {-x-6\pm \sqrt{D}\over 2}$$ where $D= -3(x-6)(x+2)>0$ for $x\in(1,2)$. So $$a_1<a<a_2$$ and now you have to find the maximum and minimum of expression(s):

$${-x-6\pm \sqrt{D}\over 2}$$ while $x\in(1,2)$.

Answer 2

Find values of $a$ such that $x^2+ax+a^2+6a<0, ∀x\in (1,2)$.

As commented by Marty Cohen, the roots must be outside the interval: $$x_1<1 \ \text{and} \ x_2>2 \iff \\ x_1=\frac{-a-\sqrt{-3a^2-24a}}{2}<1 \ \text{and} \ x_2=\frac{-a+\sqrt{-3a^2-24a}}{2}>2 \iff \\ 1) \ \sqrt{-3a^2-24a}>-a-2 \ \text{and} \ 2) \ \sqrt{-3a^2-24a}>a+4 \iff \\ 1) \begin{cases}-3a^2-24a\ge 0\\ -a-2<0\end{cases} \ \text{or} \begin{cases}-3a^2-24a\ge 0\\ a^2+7a+1<0\end{cases} \Rightarrow a\in \left(\frac{-7-3\sqrt{5}}{2},\frac{-7+3\sqrt{5}}{2}\right) \ \text{and}\\ 2) \begin{cases}-3a^2-24a\ge 0\\ a+4<0\end{cases} \ \text{or} \begin{cases}-3a^2-24a\ge 0\\ a^2+8a+4<0\end{cases} \Rightarrow a\in (-4-2\sqrt{3},-4+2\sqrt{3})\\ 1) \ \text{and} \ 2) \Rightarrow a\in \left(\frac{-7-3\sqrt{5}}{2},-4+2\sqrt{3}\right).$$ See the Desmos graph.