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This can be done using Vieta's Theorem (Described Below)

Let the solution to $x^2+3x-1$ be $k,l$. Here,

$k+l=-3$ and $kl=-1$

Other than $k,l$ there are 2 more solutions to the equation $x^4+ax^2+bx+c$, say $m,n$. 

Here,

$k+l+m+n=0 \implies m+n=-k-l=3$

$a=kl+mn+km+lm+kn+ln$

$\implies a=mn+kl+(m+n)(k+l)$

$\implies a=mn-1+3(-3)$

$\implies a=mn-10$

$-b=klm+kln+kmn+lmn$

$\implies b=kl(m+n)+mn(k+l)$

$\implies b=3(-1)-3mn$ 

$\implies b=3+3mn$

$4c=4klmn=-4mn$

So, $a+b+4c=mn-10+3mn+3-4mn=-7$

Vieta's Theorem:

Define a polynomial, $p$ as

\[p(x) = \sum_{i=0}^n c_ix^i\]

that has roots $a_1,~a_2,~a_3,~...,~a_n$. Vieta's theorem states that,

\[\sum_{i}a_i = -\frac{c_{n-1}}{c_n}\]

\[\sum_{i>j}a_ia_j = \frac{c_{n-2}}{c_n}\]

\[\sum_{i>j>k}a_ia_j = -\frac{c_{n-3}}{c_n}\]

\[\vdots\]

\[a_1a_2...a_n = (-1)^n\frac{c_0}{c_n}\]

Here, the sign alternates between positive and negative and the number of roots in each term keeps increasing. To be specific,

\[\sum_{m_1>m_2>...>m_k}a_{m_1}a_{m_2}...a_{m_k} = (-1)^k\frac{c_k}{c_n}\]