The value of $a$ for which $f(x)=x^3+3(a-7)x^2+3(a^2-9)x-1$ have a positive point of maximum 
The value of $a$ for which $f(x)=x^3+3(a-7)x^2+3(a^2-9)x-1$ have a
  positive point of maximum lies in the interval
  $(a_1,a_2)\cup(a_3,a_4)$.find the value of $a_2+11a_3+70a_4$


I found $f'(x)=3x^2+6(a-7)x+3(a^2-9)=0$ and Let the point of maximum is $x_1>0$
So $f''(x)=6x+6(a-7)<0$
$\implies 6x_1+6(a-7)<0$
But i am stuck here.The answer given in my book is $a_2+11a_3+70 a_4=320$.From this answer i cannot know what is the interval $(a_1,a_2)\cup(a_3,a_4)?$

The point of maximum is the least root of $f'(x)=0$. In order for it to be positive, both roots must be. So we need, after writing the equation as
$$
x^2+2(a-7)x+(a^2-9)=0,
$$
that the following conditions are satisfied:
\begin{cases}
-2(a-7)>0           & \text{sum of roots is positive} \\
a^2-9>0             & \text{product of roots is positive} \\
4(a-7)^2-4(a^2-9)>0 & \text{discriminant is positive}
\end{cases}
