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Need some hints how to solve this: $\sqrt{9-4\sqrt{5}}=$ ?

Thanks.

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  • $\begingroup$ roots is about the zeros of a function (see it's tag wiki). Use arithmetic for the (square) root operation instead (see the same tag wiki). Please read the tag wiki before applying a tag. $\endgroup$ – AlexR Feb 2 '15 at 15:10
  • $\begingroup$ @AlexR "radical" is an appropriate tag here, too (in lieu of "root") $\endgroup$ – Namaste Feb 2 '15 at 15:15
  • $\begingroup$ @amWhy Didn't know such an extremely localized tag existed. Thanks! $\endgroup$ – AlexR Feb 2 '15 at 15:17
  • $\begingroup$ See also: Denesting radicals at Wikipedia, Strategies to denest nested radicals (and the posts linked there). $\endgroup$ – Martin Sleziak Oct 23 '16 at 23:02
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$\sqrt{9-4\sqrt{5}}=\sqrt{5+4-2\cdot 2\cdot \sqrt{5}}=\sqrt{(\sqrt{5}-2)^{2}}=|\sqrt{5}-2|=\sqrt{5}-2$

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    $\begingroup$ I think for the sake of completeness one should mention that $\sqrt{(\sqrt5-2)^2}=|\sqrt{5}-2|$ $=\sqrt{5}-2$, so that one may not be tempted to state otherwise when it is hidden, e.g. if one wrote $$\sqrt{9-4\sqrt{5}}=\sqrt{4-2\cdot2\cdot\sqrt{5}+5}=\sqrt{(2-\sqrt{5})^2} \stackrel{\color{#F01C2C}{\rlap{\small\,\,!}{\displaystyle\triangle}}}{=} 2-\sqrt{5}.$$ $\endgroup$ – Workaholic Feb 2 '15 at 15:15
  • $\begingroup$ Didn't asked to solve this for me, but ok. I always fail on this type of arithmetic (where you need split units). Anyway, thanks. $\endgroup$ – gintko Feb 2 '15 at 15:17
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    $\begingroup$ i am sorry creatur , but i hope that u wont have problems in these type of questions anymore $\endgroup$ – avz2611 Feb 2 '15 at 15:18
  • $\begingroup$ @avz2611 Maybe you should add a small comment of how to find the completing square easily to match up for this. Nice answer, tho. (+1) $\endgroup$ – AlexR Feb 2 '15 at 15:19
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    $\begingroup$ @Workaholic How did you wizard that triangle there? :D EDIT: nvm $\color{#F01C2C}{\rlap{\small\,\,!}{\displaystyle\triangle}}$ \color{#F01C2C}{\rlap{\small\,\,!}{\displaystyle\triangle}} $\endgroup$ – AlexR Feb 2 '15 at 15:19
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This can be computed by a Simple Denesting Rule:

Here $\ 9-4\sqrt 5\ $ has norm $= 1.\:$ $\rm\ \color{blue}{subtracting\ out}\,\ \sqrt{norm}\ = 1\,\ $ yields $\,\ 8-4\sqrt 5\:$

which has $\, {\rm\ \sqrt{trace}}\, =\, \sqrt{16}\, =\, 4.\ \ \rm \color{brown}{Dividing\ it\ out}\ $ of the above yields $\ \ 2-\sqrt 5$

Remark $\ $ Many more worked examples are in prior posts on this denesting rule.

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  • $\begingroup$ Note: generally, as here, you may need to change the sign if you want the positive branch of sqrt. $\endgroup$ – Bill Dubuque Feb 2 '15 at 15:38
  • $\begingroup$ Very interesting. I didn't even know about this simple rule! $\endgroup$ – rubik Feb 2 '15 at 16:26

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