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 $\sqrt{{\mathrm{sin}}x{\mathrm{cos}}y}$, vertically $\sqrt{\frac{\frac{1}{2}}{\frac{3}{4}}}$ and $\sqrt{{{\mathrm{det}}\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)}^{2}}$, as well as $\sqrt[xyzw]{2}$, $\sqrt[\frac{\frac{1}{2}}{\frac{3}{4}}]{2}$, and $\sqrt[\lceil {det}\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)\rceil ]{2}$.

## displays

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 $$\sqrt{{\mathrm{sin}}x{\mathrm{cos}}y}$$vertically $$\sqrt{\frac{\frac{1}{2}}{\frac{3}{4}}}$$ and $$\sqrt{{{\mathrm{det}}\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)}^{2}}$$ 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 $${f}_{n}=\frac{1}{\sqrt{5}}\left[{\left(\frac{1+\sqrt{5}}{2}\right)}^{n}-{\left(\frac{1-\sqrt{5}}{2}\right)}^{n}\right]$$

## Document Tags and Contributors

**Last updated by:**fscholz,