/sci/ - Science & Math

[Anonymous No.1821718]

Hi, /sci/
Who will crack that one?

[Anonymous No.1821728]

2*sqrt(2)+1

[Anonymous No.1821738]

Wrong

[Anonymous No.1821765]

3+sqrt(2)

[Anonymous No.1821770]

3.82842

[Anonymous No.1821774]

3.82842712

[Anonymous No.1821781]

3+sqrt(2) is wrong.

[Anonymous No.1821782]

sqrt( 2+ sqrt( 9+ 4*sqrt(2)))+ sqrt(2)

A= 9+ 4*sqrt(2)= 1^2 + 2*(2*sqrt2)*1 + (2*sqrt2)^2= ( 1+ 2*sqrt2)^2

sqrt(A)= abs( 1+ 2*sqrt2)= 1+ 2*sqrt2

B= 2+ 1+ 2*sqrt2= 1^2+ 2*sqrt2*1+ (sqrt2)^2= (1+ sqrt2)^2

sqrt(B)= abs( 1+ sqrt2)= 1+ sqrt2

Ans: 1+ 2*sqrt2

[Anonymous No.1821786]

wolfram alpha knows. i don't know how he knows though.

[Anonymous No.1821799]

OP is a dumbass.

[Anonymous No.1821796]

3.82842712474619009760337744841939615713934375075389614635335947598146495692421407770077506865528314
5470027692461824594049849672111701474425288242994199871662826445331855011185511599901002305564121142
9402191199432119405490691937240294570348372817783972191046584609686174286429016795252072559905028159
7937450679309266361765928124123051670479010949150057551992345967115044067506371402270874920681699769
4320773799941398009630061088055580632908495646136985873837243161156926223193337426026031237137974474
5

i assume you see the pattern op

[Anonymous No.1821802]

http://www.wolframalpha.com/input/?i=sqrt%28sqrt%284sqrt%282%29%2B9%29%2B2%29%2Bsqrt%282

This guy

[Anonymous No.1821852]

<div class="math"> \sqrt{ 2+\sqrt{ 9+4 \sqrt{2}}} = \sqrt{2 + \sqrt{ \left( 1 + 2 \sqrt{2} \right)^2 } } </div>
<div class="math"> = \sqrt{2 + 1 + 2 \sqrt{2}} = \sqrt{ 3 + 2 \sqrt{2} } = \sqrt \left( 1 + \sqrt{2} \right)^2} = 1 + \sqrt{2} </div>

therefore
<div class="math">\sqrt{ 2+\sqrt{ 9+4 \sqrt{2}}}+sqt{2} = 1+\sqrt{2} </div>

[Anonymous No.1821857]

<div class="math"> \sqrt{ 2+\sqrt{ 9+4 \sqrt{2}}} = \sqrt{2 + \sqrt{ \left( 1 + 2 \sqrt{2} \right)^2 } } </div>

<div class="math"> = \sqrt{2 + 1 + 2 \sqrt{2}} = \sqrt{ 3 + 2 \sqrt{2} } = \sqrt{ \left( 1 + \sqrt{2} \right)^2} = 1 + \sqrt{2} </div>

therefore
<div class="math">\sqrt{ 2+\sqrt{ 9+4 \sqrt{2}}}+\sqrt{2} = 1+\sqrt{2} </div>

[Anonymous No.1821859]

<div class="math"> \sqrt{ 2+\sqrt{ 9+4 \sqrt{2}}} = \sqrt{2 + \sqrt{ \left( 1 + 2 \sqrt{2} \right)^2 } } </div>

<div class="math"> = \sqrt{2 + 1 + 2 \sqrt{2}} = \sqrt{ 3 + 2 \sqrt{2} } = \sqrt{ \left( 1 + \sqrt{2} \right)^2} = 1 + \sqrt{2} </div>

therefore
<div class="math">\sqrt{ 2+\sqrt{ 9+4 \sqrt{2}}}+\sqrt{2} = 1+2 \sqrt{2} </div>

[Anonymous No.1821896]

(2 + (9 + 4(2)^(1/2))^(1/2))^(1/2) + 2^(1/2)


Just multiply and distribute the exponents as necessary and end up with 2(2)^(1/2) + 1 aka 2sqrt(2)+1

[Anonymous No.1821905]

>>1821896
>Just multiply and distribute the exponents

Demonstrate...

[Anonymous No.1821910]

>>1821859
Thanks. How did rewriting 9+4sqrt[2] as (1 + 2sqrt[2])^2 occur to you?

[Anonymous No.1821914]

>>1821910
I needed the expression to be a square of something to cancel out the square root, so I just worked it out:

<div class="math"> 9 + 4 \sqrt{2} = (x+y \sqrt{2})^2 </div>

compare coefficients and a lil bit of grunge, and it comes out as:

<div class="math"> x=1, y=2 </div>

Now you can kill off on of the square root signs.

[Anonymous No.1821984]

Bump.