<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] We have:
$$(x^2+3x-1)(x^2+px+q)=x^4+ax^2+bx+c$$
We can expand the left side:
$$x^4+x^3(3+p)+x^2(q+3p-1)+x(3q-p)-q=x^4+ax^2+bx+c$$
Because there is no $x^3$ element on the right side, we conclude
$$3+p=0$$
$$p=-3$$
We have then:
$$x^4+x^2(q-10)+x(3q+3)-q=x^4+ax^2+bx+c$$
thus:
$$\begin{cases}a=q-10 \\ b=3q+3 \\ c=-q\end{cases}$$
Subtracting these into $a+b+4c+100$ we obtain:
$$a+b+4c+100=q-10+3q+3-4q+100 = -7+100=93$$
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