/sci/ - Science & Math » Thread #11549658

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Anonymous Fri 10 Apr 00:28:08 2020 No.11549658 View Reply Original Report

Quoted By: >>11549712 >>11549716 >>11549750 >>11549916 >>11549976 >>11549983 >>11550922

how should i calculate it?

Anonymous

Anonymous Fri 10 Apr 2020 00:37:29 No.11549688 Report

Quoted By:

just wait, its almost infinity now.

Anonymous

Anonymous Fri 10 Apr 2020 00:44:35 No.11549712 Report

Quoted By:

>>11549658
3-1/(x/8)=x

Anonymous

Anonymous Fri 10 Apr 2020 00:45:14 No.11549716 Report

Quoted By:

>>11549658
>how should i calculate it?
carefully

Anonymous

Anonymous Fri 10 Apr 2020 00:54:36 No.11549750 Report

Quoted By: >>11549770 >>11549987

>>11549658
8/(3-x)=x
8=3x-x^2
-x^2+3x-8=0
x= (-3 +- rt(9-4(-1)(-1)))/2(-1)
x= (-3 +- rt(5))/-2
x = 3+rt(5)/2

Anonymous

Anonymous Fri 10 Apr 2020 01:00:02 No.11549770 Report

Quoted By: >>11549827 >>11549935

>>11549750
but it gaves 2 answers while there is clearly 1

Anonymous

Anonymous Fri 10 Apr 2020 01:19:56 No.11549827 Report

Quoted By: >>11549928

>>11549770
How do you know?

Anonymous

Anonymous Fri 10 Apr 2020 01:50:53 No.11549916 Report

Quoted By:

>>11549658
Put it into a calculator retard

Anonymous

Anonymous Fri 10 Apr 2020 01:58:20 No.11549928 Report

Quoted By:

>>11549827
are you serious?

Anonymous

Anonymous Fri 10 Apr 2020 02:00:57 No.11549935 Report

Quoted By:

>>11549770
You factor out the negative from the bottom and top thus giving you one answer

Anonymous

Anonymous Fri 10 Apr 2020 02:22:25 No.11549976 Report

Quoted By:

>>11549658
$\displaystyle \dfrac{8}{3-x} = x \\ 8= 3x-x^2 \\ x^2-3x+8=0 \\ x= \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} = \frac{3}{2} \pm\frac{\sqrt{23}}{2} \text{i} [/marh]$

Anonymous

Anonymous Fri 10 Apr 2020 02:24:43 No.11549983 Report

Quoted By: >>11549994 >>11550922

Anonymous

Anonymous Fri 10 Apr 2020 02:25:46 No.11549987 Report

Quoted By:

>>11549750
Oh shit I forgot the 8... I fucking hate highschool garbage

Anonymous

Anonymous Fri 10 Apr 2020 02:30:49 No.11549994 Report

Quoted By: >>11551227

>>11549983
https://www.google.com/search?q=y%3Dx%5E2-3*x%2B8

Anonymous

Anonymous Fri 10 Apr 2020 03:10:55 No.11550059 Report

Quoted By: >>11551227

$\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-\frac{8}{3-}}}}}}}}}}}}}}}}}}}}}}}}$

Anonymous

Anonymous Fri 10 Apr 2020 12:30:12 No.11550922 Report

Quoted By: >>11551227

>>11549658
you can't. it doesn't converge. so we have a sequence of real numbers given by $a_1=8$ and $a_{n+1}= \frac{8}{3-a_n}$ .
this sequence doesn't converge. why?
if it did, then its limit would be a zero of a polynomial that doesn't have real zeros. see
>>11549983

Anonymous

Anonymous Fri 10 Apr 2020 15:05:38 No.11551227 Report

Quoted By:

>>11549994
>https://www.google.com/search?q=y%3Dx%5E2-3*x%2B8
Yeah its complex that's why you don't see it on the graph

>>11550059
>>11550922
The only 2 intelligent people on the thread

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