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Some people call the expressions of the form\, $a\!+\!b\sqrt{c}$\, the \PMlinkescapetext{{\em square root binomials}}, especially when $c$ is an 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 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}
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Some people call the expressions of the form\, $a+b\sqrt{c}$\, the \PMlinkescapetext{{\em square root binomials}}, especially when $c$ is an 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, 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 \PMlinkescapetext{formulae}
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| $$\sqrt{a+\sqrt{b}} = \sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$ |
$$\sqrt{a+\sqrt{b}} = \sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$ |
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and |
| $$\sqrt{a-\sqrt{b}} = \sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}}.$$ |
$$\sqrt{a-\sqrt{b}} = \sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}}.$$ |
| 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. |
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. |
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| For example, because\, $6^2\!-\!20 = 16 = 4^2$,\, we obtain |
For example, because\, $6^2\!-\!20 = 16 = 4^2$,\, we obtain |
| $$\sqrt{6\!+\!2\sqrt{5}} = \sqrt{6\!+\!\sqrt{20}} = |
$$\sqrt{6\!+\!2\sqrt{5}} = \sqrt{6\!+\!\sqrt{20}} = |
| \sqrt{\frac{6\!+\!4}{2}}+\sqrt{\frac{6\!-\!4}{2}} = 1\!+\!\sqrt{5},$$ |
\sqrt{\frac{6\!+\!4}{2}}+\sqrt{\frac{6\!-\!4}{2}} = 1\!+\!\sqrt{5},$$ |
| and because\, $4^2\!-\!7 = 9 = 3^2$,\, we get |
and because\, $4^2\!-\!7 = 9 = 3^2$,\, we get |
| $$\sqrt{4\!-\!\sqrt{7}} = \sqrt{\frac{4\!+\!3}{2}}-\sqrt{\frac{4\!-\!3}{2}} |
$$\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}.$$ |
= \frac{\sqrt{7}\!-\!1}{\sqrt{2}} = \frac{\sqrt{14}\!-\!\sqrt{2}}{2}.$$ |
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| \begin{thebibliography}{9} |
\begin{thebibliography}{9} |
| \bibitem{VA}{\sc K. V\"ais\"al\"a:} {\em Algebran oppi- ja esimerkkikirja} I. \, -- Werner S\"oderstr\"om osakeyhti\"o, Porvoo \& Helsinki (1952). |
\bibitem{VA}{\sc K. V\"ais\"al\"a:} {\em Algebran oppi- ja esimerkkikirja} I. \, -- Werner S\"oderstr\"om osakeyhti\"o, Porvoo \& Helsinki (1952). |
| \end{thebibliography} |
\end{thebibliography} |
