## BMW Coding Tool Pack Utorrent

BMW Coding Tool Pack Utorrent

BMW Coding Tool Pack Utorrent

BMW Coding Tool Pack utorrent
BMW Coding Tool Pack utorrentQ:

Solve $x^2+y^2+z^2+16x+1=0$ with the condition $(x,y,z)$ lie on the plane $x+y+z=0$

My attempt:
We can view $x^2+y^2+z^2+16x+1$ as a function of $x+y+z$ as $$\left(x+y+z\right)^2-x^2-y^2-z^2+16\left(x+y+z\right)+1$$
so we are looking for the roots of $$f(a)=a^2-a^2+16a-1$$
We now have $$a^2-a^2+16a-1=(a-4)(a-1)$$
Can I use Vieta’s formulas?

A:

Vieta’s formulas are needed to obtain a full solution. I guess this question really want to
solve the equation
$$x^2+y^2+z^2+16x+1=0$$
in parametric form $(x,y,z)=(a,b,c)$ with
$$x+y+z=0\quad \text{and} \quad a+b+c=0$$
or
$$x+y+z=0\quad \text{and} \quad a+b+c=a+b+c=-6$$
We have
$$f(a,b,c)=(a+b+c)^2-a^2-b^2-c^2+16(a+b+c)+1=$$
$$=(a^2+b^2+c^2-2abc)+(a+b+c)^2-1=0.$$
From Vieta’s formulas for $a+b+c=0$ we have
$$(a+b+c)=\frac{4a+4b+4c-3}{5}=0,$$
$$a+b+c=-6.$$
$$abc=(-6)^3=(-36),$$
a^2+b^2+c^2=\

0644bf28c6