[event_staff_9]

i am bumping this thread in case anyone needs help and doesnt know where this thread is, etc

[event_staff_9]

i am again bumping this thread for the welfare of the M2 community :D

[event_staff_9]

im back from college visiting so ask away; im leaving again Thursday afternooon

[event_staff_9]

thanks for making my life so easy you guys :D really appreciate it! :)

[Maxamillien]

I cant understand "constant term"
I got exam on Tuesday and he'll ask this for sure . .

or m=1/4 how i can find "constant term"

[spectra]

when you say constant term, do you mean value of x because after that you say m =1/4?

if so you just plug in x and solve for m

if m=5/2
(2x + 5/2x)^6 = 0
2x + 5/2x = 0
4x^2 + 5 = 0
dont have a calculator on me but do quadratic

if x = 5/2
(5 + 2m/5)^6
(5 + 2m/5)(5 + 2m/5)(5 + 2m/5)(5 + 2m/5)(5 + 2m/5)(5 + 2m/5) = 0
5 + 2m/5 = 0
2m/5 = -5
2m = -25
m = -25/2

can you explain what's being solved for though

[Maxamillien]

Originally Posted by spectra

m = 25/2
can you explain what's being solved for though

wrong...

its 4 from trot

[spectra]

(2x + m(1/x))^6
(2(5/2) + m(1/(5/2)))^6
(5 + m(2/5))^6

5 + 2m/5 = 0
25 + 2m = 0
2m = -25
m = -25/2

???

[Maxamillien]

Originally Posted by spectra

when you say constant term, do you mean value of x because after that you say m =1/4?

if so you just plug in x and solve for m

if m=5/2
(2x + 5/2x)^6 = 0
2x + 5/2x = 0
4x^2 + 5 = 0
dont have a calculator on me but do quadratic

if x = 5/2
(5 + 2m/5)^6
(5 + 2m/5)(5 + 2m/5)(5 + 2m/5)(5 + 2m/5)(5 + 2m/5)(5 + 2m/5) = 0
5 + 2m/5 = 0
2m/5 = -5
2m = -25
m = -25/2

can you explain what's being solved for though

no x not a 5/2 constant term is 5/2. thats not a quick equation. im not stupid i can do

you know combination ?

look at this:

[spectra]

i understand permutations and combinations but this doesn't relate to probability