<s>[INST] 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? 
 [/INST] 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 $q$. 
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