For $\alpha\in\mathbb{R}$, let $q(x_1, x_2) = x_1^2 + 2\alpha x_1x_2 + \dfrac{1}{2}2x_2^2$, for $(x_1, x_2) \in \mathbb{R^2}$.Find all values of $\alpha$ for which the signature of $q$ is 1.
The matrix associated with the quadratic form is $\begin{bmatrix} 1& \alpha\\\alpha&\frac12\end{bmatrix}$. To find the signature we reduce it to a diagonal form by simultaneous row and column transformations as $$\begin{bmatrix} 1& \alpha\\\alpha&\frac12\end{bmatrix}\xrightarrow[C_2\rightarrow C_2-\alpha C_1]{R_2\rightarrow R_2-\alpha R_1}\begin{bmatrix} 1& 0\\0&\frac12-\alpha^2\end{bmatrix}.$$
Signature of a matrix is defined as number of positive entries - total no of negative entries.
Signature of matrix of $q$ is $1$ if $\{\displaystyle\alpha\geq\frac{1}{2}\}\cup\{\alpha\leq\frac{-1}{2}\} $. But there is also another formula so to say viz $2p-r$ where $p$ is the number of positive entries and $r$ is the rank of the diagonal form. For $2p-r=1$ we require $p=1$ and $r=1$. Thus the values of $\alpha=\pm\dfrac{1}{2}$. Is it weird that we are getting two ranges for $\alpha$ for the same quadratic form? I am sure I have made a mistake. Can anyone please clarify?
$r$ isn't the rank of the matrix. It's the rank of a quadratic form. If the matrix is non-degenerate and $\frac 1 2 - \alpha^2 < 0$ then its signature is $1$.
For definitions see here.