<s>[INST] Rewriting $1/(v^2-(mg/k))$ as two fractions I'm looking at the solution someone gave to me and I'm having a bit of trouble following one of the steps. This step in particular,
$\frac{1}{k}\frac{1}{v^2-\frac{mg}{k}}=\frac{1}{2k\sqrt{mg/k}}(\frac{1}{v-\sqrt{mg/k}}-
\frac{1}{v+\sqrt{mg/k}})$. All letters are constants except for $v$. Any help is appreciated.
 [/INST] $$\frac{1}{k}\frac{1}{v^2-\frac{mg}{k}}=\frac{1}{k}\frac{1}{v+\sqrt\frac{mg}{k}}\frac{1}{v-\sqrt\frac{mg}{k}}=$$
$$\ =\frac{1}{k}(\frac{A}{v+\sqrt\frac{mg}{k}}+\frac{B}{v-\sqrt\frac{mg}{k}})$$
$$\text{now you bring every term in the brackets under a common denominator}$$
$$\text{so you get:}\begin{cases}
A+B=0\\
-A+B= \sqrt\frac{k}{mg}
\end{cases} \iff \begin{cases}
A=-\frac{1}{2\sqrt\frac{mg}{k}}\\
B= \frac{1}{2\sqrt\frac{mg}{k}}
\end{cases} $$
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