<s>[INST] Find the equation whose roots are each six more than the roots of $x^2 + 8x - 1 = 0$ 
Find the equation whose roots are each six more than the roots of $x^2 + 8x - 1 = 0$

I must use Vieta's formulas in my solution since that is the lesson we are covering with our teacher.
My solution:
Let p and q be the roots of the quadratic.
$$\begin{align} p + q = & -8 \\ pq = & -1 \end{align}$$
If the roots are each six more than the roots of the quadratic, then we will have:
$$\begin{align} p + q + 12 = & -8 \\ 
p + q = & -20 \\
pq = & - 3 \end{align}$$
Also, 
$$ \begin{align} (p+6)(q+6) = & pq + 6p + 6q + 36\\
=&pq + 6(p+q) + 36 \\
=&-3 + 6(-8) + 36 \\
=&-3 - 48 + 36 \\
&36 - 51 = -15 \end{align} $$
Hence, -15 = constant.
Thus, the quadratic equation is  $x^2 + 20x - 15$
My worksheet gives an answer of $x^2 - 4x - 13$, how am I wrong?
Thanks!
 [/INST] Let $r=p+6$ and $s=q+6$ be the roots of the quadratic you're trying to find.  That quadratic's coefficients are $-(r+s)$ and $rs$.  But
$$\begin{align}
r+s&=(p+6)+(q+6)\\
&=p+q+12\\
&=-8+12\\
&=4
\end{align}$$
and
$$\begin{align}
rs&=(p+6)(q+6)\\
&=pq+6(p+q)+36\\
&=-1+6(-8)+36\\
&=-1-48+36\\
&=-13
\end{align}$$
so the quadratic you're looking for is
$$x^2-(r+s)x+rs=x^2-4x-13$$
</s>