AudiForums.com

AudiForums.com (https://www.audiforums.com/forum/)
-   Off Topic (https://www.audiforums.com/forum/off-topic-5/)
-   -   If you know your math i need help (https://www.audiforums.com/forum/off-topic-5/if-you-know-your-math-i-need-help-116582/)

2.7ta6 Oct 23, 2008 06:56 PM

If you know your math i need help
 
Find all real roots of:

a. t(x) = 2x^2 -3x -6

b. f(x) = 3x^6+x^3 - 2


Evaluate

-1 - 2i
over or divided by
-5 - 4i

theTTkid Oct 23, 2008 07:21 PM

you need to do the quadratic equation

and then the bottom im pretty sure you multiply by the reciprical of the denominator so -5+4i and then i^2 = -1

MikieTaps Oct 23, 2008 07:43 PM

i am so glad i am done with school... well until i go back for my MBA...

2.7ta6 Oct 23, 2008 09:57 PM

mikietaps best cig evar.

hiwords Oct 23, 2008 10:14 PM

a) x= (3+sqrt(57))/4
x= (3-sqrt(57))/4

b) i've taken math all the way up to differential equations where we deal a lot with polynomials and have never once had to solve anything over the 3rd power. tell your teacher they're nuts and go fuck themself.

c) (13-6i)/41

TeeterTawter Oct 23, 2008 10:56 PM

i kinda got stuck watching mikietaps sig

Bad Justin Oct 23, 2008 11:28 PM


Originally Posted by TeeterTawter (Post 976198)
i kinda got stuck watching mikietaps sig

to the second power

Jc61990 Oct 24, 2008 12:56 AM

1 Attachment(s)
Attachment 50898

i tried

you should try dividing by zero

love the booB5 Oct 24, 2008 02:03 AM


Originally Posted by hiwords (Post 976159)
a) x= (3+sqrt(57))/4
x= (3-sqrt(57))/4

b) i've taken math all the way up to differential equations where we deal a lot with polynomials and have never once had to solve anything over the 3rd power. tell your teacher they're nuts and go fuck themself.

c) (13-6i)/41

seriously. there's no need to do that shit by hand. there are numerical methods that your graphing calc can employ to do that shit way faster than you ever could by hand. your teacher can eat a bag of dicks.

edit: you can actually solve b by breaking x^6 into x^3*x^3 where you treat x^3 as a variable in the second degree polynomial form a(x^3)^2 + b(x^3) + c = f(x) and solve accordingly. I stand behind my original post, though.

DÌEGØs AÜD!CTIØN Oct 24, 2008 02:06 AM

lol wat math are you in??


All times are GMT -4. The time now is 05:29 AM.


© 2026 MH Sub I, LLC dba Internet Brands