My thoughts were:
$$ x^2+ax-a=0 -p=-ax (x1x2= -ax) \ -->\ 3x2^2=-ax $$ $q=-a (x1+x2= -a) $--> $4x= -a$
$x2= -a/4, \ 3((-a/4)^2)=-ax ,\ a(3a+16x)=0$
$a=0\ a=-16x/3 $
But i dont think this is the answer ..
Sorry for such a confusing describtion, i dont know how to use this system yet
$$ x^2 + a x - a = 0 $$
Let $x_{1,2}$ be the solutions
$$ (x-x_1)(x-x_2) = x^2 + (-x_1-x_2) x + (x_1 x_2) $$
The coefficients must match. This is where you made the mistake it should have been $3x_2^2=-a$ not $-ax$.
$$ -x_1-x_2=a\\ x_1 x_2 = -a\\ $$
Now suppose that one root is 3 times the other. Without loss of generality $x_1=3x_2$ (you can always switch names if it is $x_2=3x_1$)
$$ -4x_2=a\\ 3x_2^2=-a\\ $$
Add the two equations and you get
$$ 3x_2^2-4x_2=x_2 (3x_2-4)=0 $$
so $x_2=0$ in which case you get $a=0$ or $x_2=\frac{4}{3}$ in which case $a=-\frac{4}{3}-\frac{12}{3}=\frac{-16}{3}$.
You just made that error with the $-ax$ rather than $-a$.