Show that $$\delta(k^2) \delta[(k-q_2)^2] = \delta(k^2) \delta(k^0 - \sqrt{s}/2) \frac{1}{2\sqrt{s}},$$ where $k = (k^0, \mathbf k)$ and $s = q_2^2,$ where $q_2 = (\sqrt{s},\mathbf 0)$
Attempt: I was going to use the fact that $$\delta(f(x)) = \sum \frac{\delta(x-a_i)}{|f'(a_i)|},$$ where the $a_i$ are such that $f(a_i)=0$. In this case let $f = (k-q_2)^2 = k^2 - 2k \cdot q_2 + q_2^2$ where $\cdot$ is the inner product of two four vectors. Then using the explicit four vector expressions I get ${k^0}^2 - |\mathbf k|^2 - 2k^0 \sqrt{s} + s = 0 \Rightarrow k^0 = \sqrt{s} \pm |\mathbf k|,$ which would imply I have a term $\delta(k^0 - (\sqrt{s} + |\mathbf k|))$ but this is not matching with what is given. Can anyone help? Thanks!