<s>[INST] Different ways to express $\sqrt{a}+\sqrt{b}$ I had been thinking about this for a long time. I can’t express $\sqrt{a}+\sqrt{b}$ in ways that are useful. (I’m not wanting a formula, but just some ways to express this.) I can only think of $\sqrt{a+2\sqrt{ab}+b}$.
 [/INST] You can use the conjugate:
\begin{align*}
\sqrt{a}+\sqrt{b} &= \frac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\\\
&=\frac{a^2-b^2}{\sqrt{a}-\sqrt{b}}
\end{align*}
However, this does leave a difference of square roots in the denominator. In most cases, having a sum or difference of square roots in the numerator is less of a hassle than in the denominator. 
Most of the time, you will use this technique of multiplying by the conjugate to remove pesky expressions in the denominator of a fraction. 

It also works for complex numbers, for which multiplying by the conjugate can remove the imaginary part from a denominator:
\begin{align*}
\frac{a+bi}{c+di}&=\frac{(a+bi)(c-di)}{(c+di)(c-di)}\\\\
&=\frac{(a+bi)(c-di)}{c^2+d^2}
\end{align*}

Multiplying the entire expression by $\sqrt{ab}/\sqrt{ab}$ is interesting as well:
\begin{align*}
\sqrt{a}+\sqrt{b} &= \frac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{ab}\right)}{\sqrt{ab}}\\\\
&=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}-\sqrt{b}}
\end{align*}

If for some reason you were using a log table, this identity could be useful: 
$$\sqrt{a}+\sqrt{b} = a^{1/2} + b^{1/2} = e^{(\ln a)/2} + e^{(\ln b)/2}$$
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