/sci/ - Science & Math » Thread #12023931

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MATH EQUATION

Anonymous Wed 19 Aug 04:09:56 2020 No.12023931 View Reply Original Report

Quoted By: >>12023937 >>12023941 >>12023954 >>12024174 >>12024304

How the f do u solve this?

Anonymous

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Anonymous Wed 19 Aug 2020 04:15:41 No.12023937 Report

Quoted By:

>>12023931
like using the quadratic formula my dude

Anonymous

Anonymous Wed 19 Aug 2020 04:18:36 No.12023941 Report

Quoted By: >>12023966

>>12023931
>j

Anonymous

Anonymous Wed 19 Aug 2020 04:24:24 No.12023954 Report

Quoted By: >>12023963

>>12023931
j(790x^2-395)=20x+300
j=(20x+300)/(790x^2-395)

Anonymous

Anonymous Wed 19 Aug 2020 04:27:53 No.12023963 Report

Quoted By: >>12024149

>>12023954
funny

Anonymous

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Anonymous Wed 19 Aug 2020 04:29:11 No.12023966 Report

Quoted By:

>>12023941
This terrifies the electrical engineer

Anonymous

Anonymous Wed 19 Aug 2020 05:30:19 No.12024149 Report

Quoted By:

>>12023963
what's funny? he's right

Anonymous

Anonymous Wed 19 Aug 2020 05:44:57 No.12024174 Report

Quoted By: >>12024257 >>12024274

>>12023931
Fags here making jokes but let's make that shit a diophantine equation for all x and j that are natural numbers.

Now you can't do shit right? Bitch. That's why math is a scam degree

Anonymous

Anonymous Wed 19 Aug 2020 06:35:47 No.12024257 Report

Quoted By: >>12024274

>>12024174
>Case 1 suppose j=0:
-20x - 300 = 0
x = -15 which is not a natural number so there is no solution with j=0.

Case 2 suppose x=0:
-300 - 395j = 0
j = -300/395 which is not a natural number so there is no solution with x=0.

>Case 3 suppose x>=1 and j>=1
j790x^2 - 20x - 300 - 395j
= (jx - 1)20x + (jx^2 - 1) 300 + (x^2 - 1)395j + j75x^2
>= 75
Since 75>0 there is no solution with x>0 and j>0.

Anonymous

Anonymous Wed 19 Aug 2020 06:43:43 No.12024274 Report

Quoted By:

>>12024174
>>12024257
MATHLETS BTFO

Anonymous

Anonymous Wed 19 Aug 2020 06:53:21 No.12024304 Report

Quoted By: >>12024884

>>12023931
>j
off yourself please
as for the solution would use the quadratic formula, it wasn't pretty and I probably made an error somewhere since I'm not checking my work but I got
$x=-\frac{i}{79}\pm \frac{i}{158}\sqrt{9480i-6237}$ where i is your disgusting use of j

Anonymous

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Anonymous Wed 19 Aug 2020 11:54:48 No.12024884 Report

Quoted By: >>12024900

>>12024304
oh that's what OP meant by j?

Anonymous

Anonymous Wed 19 Aug 2020 12:04:04 No.12024900 Report

Quoted By:

>>12024884
No, it's obviously a quarternion equation where the i and k components are zero. No one would be silly enough to use j to mean i.

Anonymous

Anonymous Wed 19 Aug 2020 12:16:43 No.12024915 Report

Quoted By:

Let x = a + jb and solve 2 quadratic equations separatly.

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