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/sci/ - Science & Math

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Anonymous Fri Feb 3 22:59:15 2017 No.8650405 [Reply] [Original]

How do I solve this?

>>	Anonymous Fri Feb 3 23:00:50 2017 No.8650408 TI-89

>>	Anonymous Fri Feb 3 23:01:42 2017 No.8650410 >>8650405 You can't. Galois theory proves that this is unsolvable.

>>	Anonymous Fri Feb 3 23:02:35 2017 No.8650411 File: 55 KB, 365x214, 74a.png [View same] [iqdb] [saucenao] [google] >>8650405

>>	Anonymous Fri Feb 3 23:04:10 2017 No.8650415 File: 263 KB, 558x557, 1455657897945656.png [View same] [iqdb] [saucenao] [google] >>8650411 come on man Im already banned from pol, cant a guy learn math?

>>	Anonymous Fri Feb 3 23:04:36 2017 No.8650419 this is like algebra 2 shit come on man

>>	Anonymous Fri Feb 3 23:06:31 2017 No.8650423 >>8650405 [eqn] \sqrt{\left(\sqrt{2}+\sqrt{8}\right)^2} = \sqrt{2}+\sqrt{8} = \sqrt{2} + 2\sqrt{2} = 3\sqrt{2} [/eqn]

>>	Anonymous Fri Feb 3 23:07:04 2017 No.8650425 >>8650415 read the sticky

>>	Anonymous Fri Feb 3 23:08:02 2017 No.8650427 >>8650419 Its part of a distance formula equation my answer did not equal the books,, why does the (power of 2) not cancel out the root things on 2 and 8?

>>	Anonymous Fri Feb 3 23:13:07 2017 No.8650436 chewing for you: [math] \sqrt{x^{2}} = x^{2 \cdot \frac{1}{2}} = x^{1} = x \\ \sqrt{8} = \sqrt{2 \cdot 2 \cdot 2} = \sqrt{2^{2} \cdot 2} = 2\sqrt{2} [/math]

Help Fri Feb 3 23:14:35 2017 No.8650439

>>8650427
>falling for the bait
The square can be used to cancel the outer root. Using the root on the other two would be like multiplying (√2 + √8) × (√2 + √8) which ends up being (√2√2 + √2√8 + √8√2 + √8√8) which is equal to (2 + 2√2√8 + 8)
If you need more help just ask

>>	Anonymous Fri Feb 3 23:17:34 2017 No.8650443 >>8650436 How do I write in math?

>>	Anonymous Fri Feb 3 23:19:30 2017 No.8650449 >>8650443 read the sticky

>>	Anonymous Fri Feb 3 23:32:42 2017 No.8650471 File: 3 KB, 212x32, fff.png [View same] [iqdb] [saucenao] [google] >>8650439 I needed to find the Distance between these points the books answer is >square root of 83 I did not get this answer. How would this be calculated using the distance formula?

>>	Anonymous Fri Feb 3 23:41:57 2017 No.8650494 >>8650411 i once reported a thread in /sp/ with a picture of a naked black kid urinating on a woman and got banned for "Submitting false or misclassified reports". i will never report anything again. fuck jannies, mods, and the gook moot

Julio Profe Fri Feb 3 23:41:58 2017 No.8650495

>>8650405
Finally someone post something challenging.
It took me about 30 minutes to solve, it was tougher than I originally thought. Here's my solution:
[eqn]\sqrt{(\sqrt{2}+\sqrt{8})^2} = \sqrt{(2+2\sqrt{2}\sqrt{8} +8)=}[/eqn]
[eqn]\sqrt{10+2\sqrt{2}\sqrt{8}}= \sqrt{10+(2^1*2^{1/2}*8^{1/2})=}[/eqn]
[eqn]\sqrt{10+(2^1*2^{1/2}*(2^3)^{1/2})} = \sqrt{10+(2^1*2^{1/2}*2^{3/2})} = [/eqn]
[eqn]\sqrt{10+(2^{2/2}*2^{1/2}*2^{3/2})}=\sqrt{10+(2^{6/2})}[/eqn]
[eqn]\sqrt{10+(2^{3}})= \sqrt{10+(8})= \sqrt{18}=\sqrt{3*3*2}[/eqn]
[eqn]\sqrt{3^2*2}= (3^2*2)^{1/2}= (3^2)^{1/2}*2^{1/2}=3^{2/2}*2^{1/2}=
3^1*2^{1/2}=3\sqrt{2} [/eqn]
Thanks anon for posting real problems!!!!!

>>	Anonymous Fri Feb 3 23:47:53 2017 No.8650511 >>8650471 find the distance between x1 and x2, find the distance between y1 and y2, now you got both catheti, use pythagoras >>>/sci/sqt

>>	Anonymous Fri Feb 3 23:56:03 2017 No.8650521 >>8650443 [math]With[/math] [math]science[/math]

>>	Anonymous Fri Feb 3 23:58:11 2017 No.8650524 >>8650511 ok I got the correct answer now I had forgot to do something

>>	Anonymous Sat Feb 4 01:11:02 2017 No.8650628 >>8650495 >Julio Profe Kek

>>	Anonymous Sat Feb 4 18:38:21 2017 No.8652228 >>8650405 The answer is 3x root2

>>	Anonymous Sat Feb 4 18:41:39 2017 No.8652235 File: 1.00 MB, 2000x2000, 1443370078997.png [View same] [iqdb] [saucenao] [google] >>8650415 nope /r9k/ may work for you though

>>	Anonymous Sat Feb 4 18:53:56 2017 No.8652258 sqrt(2)+sqrt(8) Or you can perform the square and get sqrt(2+2(4)+8) = sqrt(18) = 3sqrt(2)

>>	Anonymous Sat Feb 4 19:05:55 2017 No.8652274 File: 383 KB, 800x600, 1482164421724.png [View same] [iqdb] [saucenao] [google] >all these brainlets saying sqrt(x^2)=x and not \|x\|

>>	Anonymous Sat Feb 4 19:15:38 2017 No.8652293 [math]\sqrt(\sqrt2+\sqrt8)^{2}[/math] [math]\sqrt(2+\sqrt16+8)[/math] [math]\sqrt14[/math] I felt ridiculous typing this out.

>>	Anonymous Sat Feb 4 19:18:33 2017 No.8652303 >>8652274 >[math]\displaystyle\sqrt{\left(x^2\right)}=-\|x\| : x\in\mathbb{Z^+}[/math] Yeah okay. >>8652293 It's [math]\displaystyle2\sqrt{16}[/math]

>>	Anonymous Sat Feb 4 19:19:24 2017 No.8652305 >>8652293 even more ridiculous when you got it wrong senpai

>>	Anonymous Sat Feb 4 20:18:06 2017 No.8652434 File: 96 KB, 300x181, 1445564256697.png [View same] [iqdb] [saucenao] [google] >>8650415 kek

>>	Anonymous Sat Feb 4 20:35:06 2017 No.8652471 The sqrt cancels the square. You're left with sqrt(2)+sqrt(8) Which is equal to sqrt(2)+2sqrt(2) combine terms, 3sqrt(2)

>>	Anonymous Sat Feb 4 20:36:46 2017 No.8652474 >>8650495 You're an asshole. I like u tho.

>>	Anonymous Sat Feb 4 21:28:12 2017 No.8652586 >>8650443 Admin privileges.

>>	Anonymous Sat Feb 4 21:52:58 2017 No.8652647 >>8650443 You buy a 4chan pass and the math panel will display in your post/ reply box

>>	Anonymous Sat Feb 4 21:58:01 2017 No.8652653 Isn't this just the square root of 10? What am I missing?

>>	Anonymous Sat Feb 4 22:55:57 2017 No.8652761 >>8652653 Dude I think the square root and the square cancel out right?? so it should be Root 2+ Root 8... Some dude got that in this thread except he showed his work which was unnecessary.

>>	Anonymous Sun Feb 5 00:14:34 2017 No.8652885 >>8650423 wrong [math] \displaystyle \sqrt{\left(\sqrt{2}+\sqrt{8}\right)^2} = \left \| \sqrt{2}+\sqrt{8} \right \| = \sqrt{2} + 2\sqrt{2} = 3\sqrt{2} [/math]

>>	Anonymous Sun Feb 5 00:18:17 2017 No.8652891 >>8652653 >What am I missing a brain

>>	Anonymous Sun Feb 5 00:23:15 2017 No.8652894 >>8652885 > assuming non-positive roots of positive numbers shiggy

>>	Anonymous Sun Feb 5 02:45:52 2017 No.8653004 >>8652894 >convention is assumption donatello

>>	Anonymous Sun Feb 5 03:28:20 2017 No.8653042 File: 3 KB, 367x110, x.png [View same] [iqdb] [saucenao] [google] >>8652894 https://en.wikipedia.org/wiki/Square_root

>>	Anonymous Sun Feb 5 04:45:06 2017 No.8653110 >>8653004 What? I'm assuming that the square roots are gonna give positive numbers because I see no complex numbers. [math] \sqrt{x^2}: \mathbb{R}^+ \to \mathbb{R}^+ [/math] is the identity function >>8653042 But x > 0 you tard

>>	Anonymous Sun Feb 5 04:50:57 2017 No.8653115 >>8653110 the point is to assume nothing. ABS is in the definition, use it always

>>	Anonymous Sun Feb 5 04:57:51 2017 No.8653124 >>8653115 > ABS is in the definition, use it always it's not. but w/e

>>	Anonymous Sun Feb 5 20:40:11 2017 No.8654645 >>8650521

>>	Anonymous Mon Feb 6 08:10:19 2017 No.8655459 >>8650405 just type it into your calculator or something

Anonymous Mon Feb 6 09:35:29 2017 No.8655564

>>8650405

[eqn]\sqrt{(\sqrt{2}+\sqrt{8})^{2} } [/eqn]
[eqn]= \sqrt{(\sqrt{2}+\sqrt{8})(\sqrt{2}+\sqrt{8}) } [/eqn]
[eqn]= \sqrt{(2 + \sqrt{16} + \sqrt{16} + 8)} [/eqn]
[eqn]= \sqrt{(10+4+4)} [/eqn]
[eqn]= \sqrt{18} [/eqn]
[eqn]= \sqrt{9\cdot 2} [/eqn]
[eqn]= 3\sqrt{2} [/eqn]

>>	Anonymous Mon Feb 6 13:28:45 2017 No.8655932 >>8650427 >cancel out L0Lno fgt pls

>>	Anonymous Mon Feb 6 15:50:39 2017 No.8656251 >>8655564 retard

Anonymous Mon Feb 6 16:14:53 2017 No.8656303

>>8650405
[math]
\sqrt{\left (\sqrt 2 + \sqrt 8\right ) ^2} =
\sqrt{\left (\sqrt 2 + 2 \sqrt 2\right ) ^2}
[/math]
[math]
= \sqrt {\left (3 \sqrt 2
\right ) ^2 } = \sqrt {\left ( \sqrt 18\right )^2}
[/math]
[math] = \sqrt {18} = \sqrt {2 \cdot 9} = 3 \sqrt 2
[/math]

>>	Anonymous Mon Feb 6 17:46:24 2017 No.8656514 >>8650415 How do you get banned from /pol/ I thought being a shitty board has become the core, frame and shell of /pol/ so thoroughly that the mods gave up?

>>	Anonymous Mon Feb 6 18:52:39 2017 No.8656641 >>8652235 >Weishaupt's plateau I still don't get it

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