Show that $\langle x, Ax \rangle + \langle b, x \rangle = c$ can be transformed to $\langle x', Ax' \rangle = 1$ Let $A$ be a real, regular, symmetric $n \times n$ matrix, $b \in \mathbb{R}^n$ and $c \in \mathbb{R}$
How can I show that $$\langle x, Ax \rangle + \langle b, x \rangle = c$$ can be transformed by using $$x' = \alpha x + \beta$$ where $0 \neq \alpha \in \mathbb{R}$, $\beta \in \mathbb{R}^n$ to $$\langle x', Ax' \rangle = 1$$ if 
$c + \frac{1}{4}\langle b, A^{-1}b \rangle > 0$
I tried to start with $\langle x, Ax \rangle + \langle b, x \rangle = c$ and $\langle x', Ax' \rangle = 1$ by inserting $x' = \alpha x + \beta$ but I always end up with a large equation.
Does anybody see how I can solve this problem?
Thanks a lot!
 A: You want to use "completing the square" from your algebra-2 class. 
Roughly, you've got
$$
ax^2 + bx = c,
$$
and completing the square would rewrite this as 
$$
x^2 + a^{-1}bx = a^{-1}c
$$
and then 
$$
x^2 + a^{-1}bx + (a^{-1}b)^2/4 = a^{-1}c + (a^{-1}b)^2/4
$$
after which you'd let 
$$
x' = x + (a^{-1}b)/2,
$$
and get
$$
x'^2 = a^{-1}c + (a^{-1}b)^2/4.
$$
Can you try to follow that pattern here? 
In particular, if you let 
$$
x' = x + \frac{1}{2}A^{-1}b,
$$
then 
\begin{align}
\langle x', Ax' \rangle> 
&= \langle x + \frac{1}{2}A^{-1}b, A(x + \frac{1}{2}A^{-1}b) \rangle \\
&= \langle x , A(x + \frac{1}{2}A^{-1}b) \rangle + \langle \frac{1}{2}A^{-1}b, A(x + \frac{1}{2}A^{-1}b) \rangle \\
&= \langle x , Ax \rangle + 
   \langle x, A\frac{1}{2}A^{-1}b) \rangle + 
   \langle \frac{1}{2}A^{-1}b, Ax \rangle +  
   \langle \frac{1}{2}A^{-1}b, A\frac{1}{2}A^{-1}b) \rangle \\
&= \langle x , Ax \rangle + 
   \frac{1}{2}\langle x, b) \rangle + 
   \frac{1}{2}\langle A^{-1}b, Ax \rangle +  
   \frac{1}{4}\langle A^{-1}b, b \rangle \\
&= \langle x , Ax \rangle + 
   \frac{1}{2}\langle x, b) \rangle + 
   \frac{1}{2}\langle b, x \rangle +  
   \frac{1}{4}\langle A^{-1}b, b \rangle \\
&= \langle x , Ax \rangle + 
   \langle x, b \rangle + 
   \frac{1}{4}\langle A^{-1}b, b \rangle \\
\end{align}
That seems to me as if it's getting pretty close to what you want, no? 
A: Long story short: we can rewrite the above as
$$
x^TAx + b^Tx - c = 0
$$
Now, set $x = x' - \frac 12 A^{-1}b $.  Substituting this into the above gives you
$$
0 = (x' - \frac 12 A^{-1}b)^TA(x' - \frac 12 A^{-1}b) + b^T(x' - \frac 12 A^{-1}b) - c = \\
\left[(x')^T A x' - b^T x' + \frac 12 b^TA^{-1}b\right] +
\left[ b^T x' - \frac 12 b^T A^{-1}b\right] - c =\\
(x')^T A x' - c
$$
So, setting $x'$ to $x + \frac 12 A^{-1}b$ gives us the equation
$$
(x')^T A x' = c \implies 
\left(\frac 1{\sqrt c} x'\right)^T A \left(\frac 1{\sqrt c} x'\right) = 1 \implies\\
\left\langle 
\left(\frac 1{\sqrt c} x'\right), A \left(\frac 1{\sqrt c} x'\right)
\right \rangle = 1
$$
So, setting $\tilde x = \frac 1{\sqrt{c}} x' = \frac 1 {\sqrt{c}}(x + \frac 12 A^{-1})$ means that
$$
\langle x,Ax \rangle + \langle x,b \rangle = c \iff
\langle \tilde x, A \tilde x \rangle = 1
$$
