The roots of the equation $x^2+3x-1=0$ are also the roots of $x^4+ax^2+bx+c=0$ 
The roots of the equation $x^2+3x-1=0$ are also the roots of quartic equation $x^4+ax^2+bx+c=0$. Find $a+b+4c$.

This problem is from yesterday's Bangladesh National Math Olympiad 2017. I tried this using Vieta Root Jumping but no luck. After the contest my friend laughed at me "One doesn't simply try a 10 point problem with Vieta Root Jumping".
How to solve this problem? 

For the equation $x^2+3x-1=0$, the sum of the roots is $-3$ and the product is $-1$.  For the equation $x^4+ax^2+bx+c=0$, the sum of the roots is $0$ and the product is $c$.  If the roots of the first equation are roots of the second, the remaining two roots have sum $3$ and product $-c$.  Therefore, we have
$$x^4+ax^2+bx+c=(x^2+3x-1)(x^2-3x-c)$$
The coefficient of the $x^2$ term is $-c-9-1=-c-10$.  The coefficient of $x$ is $-3c+3$.  Therefore,
$$a+b+4c=-c-10-3c+3+4c=-7$$
