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? 
 A: Note that if $x^2+3x-1=0$, then roots of the quadratic $x^2+3x-1$ are also roots of $$(x^2+3x-1)(x^2-px+q)$$
Which follows from polynomial long divison. 
Since the coefficient of $x^3$ is $0$, we have that $p=3$. Now note that $$(x^2+3x-1)(x^2-3x+q)=x^4+(q-10)x^2+(3q+3)x-q$$
So the value of $a+b+4c=q-10+3q+3-4q=-7$. So the answer is $-7$ no matter the value of $q$. 
A: $$(x^2+3x-1)(x^2-3x+3)=x^4-7x^2+12x-3$$ 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.
A: 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,
$$a+b+4c=-c-10-3c+3+4c=-7$$
