square root of square root binomialSquareRootOfSquareRootBinomial2005-06-21 07:02:122005-06-30 14:19:53Topicsquare rootsquare root binomialsquare root polynomialnested square roots% this is the default PlanetMath preamble. as your knowledge
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\newtheorem*{thmplain}{Theorem}Some people call the expressions of the form\, $a\!+\!b\sqrt{c}$\, the \PMlinkescapetext{{\em square root binomials}}, especially when $c$ is an square-free integer greater than 1 (and $a$ and $b$ rational numbers).\, On the high school \PMlinkescapetext{level one may learn to perform arithmetic operations between such binomials} (see e.g. division), or also polynomials containing several square root \PMlinkescapetext{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 \PMlinkescapetext{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 \PMlinkescapetext{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
$$\sqrt{4\!-\!\sqrt{7}} = \sqrt{\frac{4\!+\!3}{2}}-\sqrt{\frac{4\!-\!3}{2}}
= \frac{\sqrt{7}\!-\!1}{\sqrt{2}} = \frac{\sqrt{14}\!-\!\sqrt{2}}{2}.$$
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\bibitem{VA}{\sc K. V\"ais\"al\"a:} {\em Algebran oppi- ja esimerkkikirja} I. \, -- Werner S\"oderstr\"om osakeyhti\"o, Porvoo \& Helsinki (1952).
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