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? 

$$(x^2+3x-1)(x^2-3x+3)=x^4-7x^2+12x-3$$ has the same real roots as $x^2+3x-1$. 
In this case $a=-7,b=12,c=-3$ so $$a+b+4c=-7.$$
Moreover $$(x^2+3x-1)(x^2-3x+d)=x^4+(d-10)x^2+(3d+3)x-d$$ for $$d>\frac{9}{4}$$ has the same real roots as $x^2+3x-1$. In this case $a=d-10,b=3d+3,c=-d$ so $$a+b+4c=-7.$$
So if the statement of the problem says the same real roots then it seems to be that $a+b+4c=-7$ in all cases.
