Differential Geometry and Its Application - Exercise 3.1.7

Problem:

Show that the principal curvatures are given in terms of $K$ and $H$ by

$k_1 = H + \sqrt{H^2 - K}$ and $k_2 = H - \sqrt{H^2 - K}$.

Solution:

We know $H = \frac{k_1 + k_2}{2}$ and $K = k_1 k_2$. Therefore we can form the quadratic equation $x^2 - 2Hx + K$ so that the roots are $k_1$ and $k_2$

Now using the quadratic formula, we get

$k = \frac{2H \pm \sqrt{4H^2 - 4K}}{2} = H \pm \sqrt{H^2 - K}$, this is exactly what we needed.