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
 A: We can find the equation must satisfy$x^4+ax^2+bx+c=(x^2+3x-1)(x^2+mx+n)$ and$m=-3,n=-c$. So when$x=1,1+a+b+c=3(1-3-c)\rightarrow a+b+4c+7=0$.
A: Hint: The solutions of your polynomial are given by
$$x_{1,2}=-\frac{3}{2}\pm\frac{\sqrt{13}}{2}$$
plugging $x_1$ into the other equation we get
$$b\sqrt{13}+3c\sqrt{13}-3\sqrt{13}+2a+3b+11c+11=0$$
the rest is for you!
A: 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:
$$a+b+4c=-7 \Rightarrow a+b+4c+100=93.$$
A: 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:
$$a+b+4c+100=q-10+3q+3-4q+100 = -7+100=93$$
