MathML has two root objects, an <msqrt> $\sqrt{x}$ and an <mroot> $\sqrt[3]{x}$. These are pretty simple. About all you can do with them is see how the rendering stretches them in various ways: horizontally $\sqrt{\sin x\cos y}$, vertically $\sqrt{\frac{\frac{1}{2}}{\frac{3}{4}}}$ and $\sqrt{{\det \left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)}^2}$, as well as $\sqrt[xyzw]{2}$, $\sqrt[\frac{\frac{1}{2}}{\frac{3}{4}}]{2}$, and $\sqrt[\lceil det\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)\rceil ]{2}$.

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MathML has two root objects, an <msqrt> $$\sqrt{x}$$ and an <mroot> $$\sqrt[3]{x}$$These are pretty simple. About all you can do with them is see how the rendering stretches them in various ways: horizontally $$\sqrt{\sin x\cos y}$$vertically $$\sqrt{\frac{\frac{1}{2}}{\frac{3}{4}}}$$ and $$\sqrt{{\det \left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)}^2}$$ as well as $$\sqrt[xyzw]{2}$$$$\sqrt[\frac{\frac{1}{2}}{\frac{3}{4}}]{2}$$and $$\sqrt[\lceil det\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)\rceil ]{2}$$

The formula of Binet shows how the n-th term in the Fibonacci series can be expressed using roots $$f_n=\frac{1}{\sqrt{5}}\left[{\left(\frac{1+\sqrt{5}}{2}\right)}^n-{\left(\frac{1-\sqrt{5}}{2}\right)}^n\right]$$