/sci/ - Science & Math » Thread #12388937

2KiB, 176x54, sqrt 5 answer.gif

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Anonymous Thu 26 Nov 15:15:33 2020 No.12388937 View Reply Original Report

Quoted By: >>12388959 >>12389501 >>12390479 >>12390514 >>12390770

I was trying to find the square root of (5+sqrt(5)) and wolfram gave me this answer. How do you get this answer with pen and paper? Also, what do they call number that is a combination of an integer and a radical?

Anonymous

Anonymous Thu 26 Nov 2020 15:34:02 No.12388959 Report

Quoted By: >>12388971

>>12388937
>combination of an integer and a radical
No name exists.

Anonymous

Anonymous Thu 26 Nov 2020 15:39:35 No.12388971 Report

Quoted By:

>>12388959
Which is probably why I can't find information about how to do this online

Anonymous

Anonymous Thu 26 Nov 2020 17:08:15 No.12389167 Report

Quoted By: >>12390514

I guess you write $\sqrt{5+ \sqrt{5}}$ as $\sqrt{5 + \sqrt{5}} ( cos( n \pi ) + i sin(n \pi ) )$ and go from there.

Anonymous

Anonymous Thu 26 Nov 2020 18:53:55 No.12389501 Report

Quoted By: >>12390474

>>12388937
Generally problems like this are called nested radical problems and they don't always admit clean solutions.
"Most" expressions of this form can be simplified into $sqrt(a) \pm sqrt(b)$ where a and b are rationals (potentially complex rationals), but this does not seem to be one of these cases (given the solution has a 4th root).
There are a few other tricks as well, related to roots of 4th powers, but none of those seem to work cleanly on this one.
I'm curious if someone here has a good method to simplify your original expression into that form without nested radicals.

Anonymous

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Anonymous Thu 26 Nov 2020 19:46:01 No.12389659 Report

Quoted By: >>12390474 >>12390583

Anonymous

Anonymous Fri 27 Nov 2020 00:56:16 No.12390474 Report

Quoted By: >>12390760

>>12389501
>>12389659
I still have no idea what this means but thanks, I appreciate the input

Anonymous

Anonymous Fri 27 Nov 2020 00:59:29 No.12390479 Report

Quoted By:

>>12388937
What number does that equation shit out? It does give an answer right?

Anonymous

Anonymous Fri 27 Nov 2020 01:19:11 No.12390514 Report

Quoted By:

>>12388937
This is a case of computer stupidity. The expression is clearly real but it followed a procedure like >>12389167 to obtain a spectacularly unenlightening result

Anonymous

Anonymous Fri 27 Nov 2020 01:57:07 No.12390583 Report

Quoted By:

>>12389659
solution chain looks okay

checks out to 60 digits

actually impressive

Anonymous

Anonymous Fri 27 Nov 2020 03:27:54 No.12390760 Report

Quoted By:

>>12390474
(1-2i)(1+2i)=1+2i-2i-(2i*2i)=1-4(-1)=5

Anonymous

Anonymous Fri 27 Nov 2020 03:30:14 No.12390770 Report

Quoted By:

>>12388937
my calculator says the answer is 2.689994047855829307840235875543, which rounds to 3. you can thank me later

Anonymous

Anonymous Fri 27 Nov 2020 04:24:10 No.12390926 Report

Quoted By:

$ 5+ \sqrt{5}= \sqrt{5}\left ( \sqrt{5}+1 \right ) = \dfrac{ \sqrt{5}}{2} \left ( 2 \sqrt{5}+2 \right ) = \dfrac{ \sqrt{5}}{2} \left ( 2 \sqrt{5}+1+1 +2i-2i \right ) = \dfrac{ \sqrt{5}}{2} \left ( 1+2i +2 \sqrt{5} +1-2i \right ) = \dfrac{ \sqrt{5}}{2} \left ( \sqrt{1+2i}^{2} +2 \sqrt{5} +\sqrt{1-2i}^{2} \right ) = \dfrac{ \sqrt{5}}{2} \left ( \sqrt{1+2i}^{2} +2 \sqrt{(2+i)(2-i)} +\sqrt{1-2i}^{2} \right ) = \dfrac{ \sqrt{5}}{2} \left ( \sqrt{1+2i}^{2} +2 \sqrt{(2+i)(2-i)} +\sqrt{1-2i}^{2} \right ) = \dfrac{ \sqrt{5}}{2} \left ( \sqrt{1+2i}^{2} +2 \sqrt{2+i} \sqrt{2-i} +\sqrt{1-2i}^{2} \right ) \\ = \dfrac{ \sqrt{5}}{2} \left ( \sqrt{2+i} + \sqrt{2-i} \right )^{2} \\ \Rightarrow \sqrt{5+ \sqrt{5}} = \dfrac{ \sqrt[4]{5}}{\sqrt{2}} \left ( \sqrt{2+i} + \sqrt{2-i} \right ) $

Anonymous

Anonymous Fri 27 Nov 2020 04:25:51 No.12390934 Report

Quoted By:

$ 5+ \sqrt{5}= \sqrt{5}\left ( \sqrt{5}+1 \right ) = \dfrac{ \sqrt{5}}{2} \left ( 2 \sqrt{5}+2 \right ) = \dfrac{ \sqrt{5}}{2} \left ( 2 \sqrt{5}+1+1 +2i-2i \right ) = \dfrac{ \sqrt{5}}{2} \left ( 1+2i +2 \sqrt{5} +1-2i \right ) = \dfrac{ \sqrt{5}}{2} \left ( \sqrt{1+2i}^{2} +2 \sqrt{5} +\sqrt{1-2i}^{2} \right ) \\ = \dfrac{ \sqrt{5}}{2} \left ( \sqrt{1+2i}^{2} +2 \sqrt{(2+i)(2-i)} +\sqrt{1-2i}^{2} \right ) = \dfrac{ \sqrt{5}}{2} \left ( \sqrt{1+2i}^{2} +2 \sqrt{(2+i)(2-i)} +\sqrt{1-2i}^{2} \right ) = \dfrac{ \sqrt{5}}{2} \left ( \sqrt{1+2i}^{2} +2 \sqrt{2+i} \sqrt{2-i} +\sqrt{1-2i}^{2} \right ) \\ = \dfrac{ \sqrt{5}}{2} \left ( \sqrt{2+i} + \sqrt{2-i} \right )^{2} \\ \Rightarrow \sqrt{5+ \sqrt{5}} = \dfrac{ \sqrt[4]{5}}{\sqrt{2}} \left ( \sqrt{2+i} + \sqrt{2-i} \right ) $

Anonymous

Anonymous Fri 27 Nov 2020 04:28:50 No.12390945 Report

Quoted By:

$ 5+ \sqrt{5}= \sqrt{5}\left ( \sqrt{5}+1 \right ) = \dfrac{ \sqrt{5}}{2} \left ( 2 \sqrt{5}+2 \right ) = \dfrac{ \sqrt{5}}{2} \left ( 2 \sqrt{5}+1+1 +2i-2i \right ) = \dfrac{ \sqrt{5}}{2} \left ( 1+2i +2 \sqrt{5} +1-2i \right ) = \dfrac{ \sqrt{5}}{2} \left ( \sqrt{1+2i}^{2} +2 \sqrt{5} +\sqrt{1-2i}^{2} \right ) \\ = \dfrac{ \sqrt{5}}{2} \left ( \sqrt{1+2i}^{2} +2 \sqrt{(2+i)(2-i)} +\sqrt{1-2i}^{2} \right ) = \dfrac{ \sqrt{5}}{2} \left ( \sqrt{1+2i}^{2} +2 \sqrt{2+i} \sqrt{2-i} +\sqrt{1-2i}^{2} \right ) \\ = \dfrac{ \sqrt{5}}{2} \left ( \sqrt{2+i} + \sqrt{2-i} \right )^{2} \\ \Rightarrow \sqrt{5+ \sqrt{5}} = \dfrac{ \sqrt[4]{5}}{\sqrt{2}} \left ( \sqrt{2+i} + \sqrt{2-i} \right ) $

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