Some people call the expressions of the form $a\!+\!b\sqrt{c}$ , the square root binomials, especially when $c$ is an square-free integer greater than 1 (and $a$ and $b$ rational numbers). On the high school level one may learn to perform arithmetic operations between such binomials (see e.g. division), or also polynomials containing several square root terms. Taking the square root of a square root binomial is more difficult and usually results nested square roots. However, there are some exceptions if the numbers are appropriate. We have the formulae $$\sqrt{a+\sqrt{b}} = \sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$ and $$\sqrt{a-\sqrt{b}} = \sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}}.$$ If $a^2\!-\!b$ , happens to be square of a rational number, then the formulae allow to convert the square roots on the left side into expressions without nested square roots.

For example, because $6^2\!-\!20 = 16 = 4^2$ we obtain $$\sqrt{6\!+\!2\sqrt{5}} = \sqrt{6\!+\!\sqrt{20}} = \sqrt{\frac{6\!+\!4}{2}}+\sqrt{\frac{6\!-\!4}{2}} = 1\!+\!\sqrt{5},$$ and because $4^2\!-\!7 = 9 = 3^2$ we get $$\sqrt{4\!-\!\sqrt{7}} = \sqrt{\frac{4\!+\!3}{2}}-\sqrt{\frac{4\!-\!3}{2}} = \frac{\sqrt{7}\!-\!1}{\sqrt{2}} = \frac{\sqrt{14}\!-\!\sqrt{2}}{2}.$$

Bibliography

1

K. V¨AISÄLÄ: Algebran oppi- ja esimerkkikirja I. - Werner Söderström osakeyhtiö, Porvoo & Helsinki (1952).