<s>[INST] If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$ 
If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$

I tried really hard but the most I could get is the sum of the roots of the second equation is $3$
Please could someone solve this please !
It would mean the world to me
 [/INST] Alternatively, the Vieta's formula's for $x^2 + 3x -1 = 0$:
$$x_1+x_2=-3, x_1x_2=-1 \Rightarrow x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=9+2=11.$$
Plugging $x_1,x_2$ to $x^4 + ax^2 + bx + c = 0$ and subtracting:
$$\begin{cases}x_1^4 + ax_1^2 + bx_1 + c = 0\\ x_2^4 + ax_2^2 + bx_2 + c = 0\end{cases} \Rightarrow \\
(x_1-x_2)(x_1+x_2)(x_1^2+x_2^2)+a(x_1-x_2)(x_1+x_2)+b(x_1-x_2)=0 \Rightarrow \\
\require{cancel}\cancel{(x_1-x_2)}[(-3)(11)+a(-3)+b]=0 \Rightarrow \\ 3a-b=-33 \ \ \ \ (1)$$
and adding:
$$(x_1^2+x_2^2)^2-2(x_1x_2)^2+a(x_1^2+x_2^2)+b(x_1+x_2)+2c=0 \Rightarrow \\
11^2-2+11a-3b+2c=0 \Rightarrow \\
119+2a+3(\underbrace{3a-b}_{=-33})+2c=0 \Rightarrow \\
a+c=-10 \ \ \ \ (2)$$
Multiply $(2)$ by $4$ and subtract $(1)$ to get:
$$a+b+4c=-7 \Rightarrow a+b+4c+100=93.$$
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