Completing the square to find if quadratic form is positive definite. I have the quadratic form
$$g=x_1^2+6x_2^2+8x_3^2-4x_1x_2-6x_1x_3-x_2x_3$$
I have problems completing the square. I tried to rewrite the expression as follows
$$g=x_1^2-4x_1x_2+6x_2^2-x_2x_3+8x_3^2$$ 
Hence, 
$$((x_1-2{x_2})^2 -(2x_2)^2) + 6 \left( x_2-\frac{x_3}{12} \right)^2 - \left( \frac{x_3}{2} \right)^2 + 8x_3^2$$
Can I do it like this? On complete square calculator, it's said that you cannot complete the square for expression like this. Anyway, the reason I need to complete the square for this expression is because I need to determine whether the given quadratic form is positive definite. But I don't know how to proceed.
 A: Construct the coefficient matrix
$$A = \begin{bmatrix} 1& -2& -3\\ -2& 6& -1/2 \\-3 &-1/2 & 8\end{bmatrix}$$
Find the determinants
$D_1 = |a_{11}| = 1 > 0$
$D_2 = \begin{vmatrix}1&-2 \\ -2&6\end{vmatrix} = 6 -4 =2 > 0$
$D_3 = \det(A) = -177/4 < 0$

So, the given form is indefinite in nature.

For an nxn matrix if  

$\bullet D_1 , D_2, \cdots D_n>0 \implies$ Positive definite 
$\bullet D_1,D_3,D_5\cdots<0 $ and $D_2,D_4,D_6\cdots>0$ or $(-1)^kD_k >0   , \  k =0,1,\cdots n \implies$ Negative definite  
$\bullet D_1 , D_2, \cdots D_n \ge0$ or $D_k>0 $  with at least one value zero $\implies$ Positive semi-definite
$\bullet(-1)^kD_k \ge0   , \  k =0,1,\cdots n$ with at least one zero $\implies$ Negative semi-definite  
$\bullet$All other cases are indefinite

A: Note that if $x_2=0$ then the quadratic form $g$ is 
$$x_1^2+8x_3^2-6x_1x_3=(x_1-2x_3)(x_1-4x_3)$$ 
whose sign is indefinite. So $g$ is NOT positive definite.
A: Completing the square (Lagrange's Method):
$$x^2+6y^2+8z^2-4xy-6xz-yz=\left(x-(2y+3z)\right)^2-4y^2-12yz-9z^2+6y^2+8z^2-yz$$
$$=\left(x-(2y+3z)\right)^2+2y^2-13yz-z^2=\left(x-(2y+3z)\right)^2+2\left(y-\frac{13}4z\right)^2-\frac{177}8z^2$$
Thus, for example, $\;g(38,\,13,\,4)=-177\cdot2=-354<0\;$ and thus the form isn't positive definite.
BTW, $\;g(1,0,0)=1\;$ , so the form is actually indefinite.
