anyone good at algebra?

[bryantm3]

someone show me how to do this. i don't want just the answer.

X to the power of 2/3, minus 9x to the power of 1/3, plus eight equals zero.

[jOHN rODRIGUEZ]

You show us how you think it through first. :D

[Dunwho]

Oh man ... i had a sit down and gave it a go, i havnt done maths since leaving school...

...im actually going to try and figure it out properly... but something is telling me to find two different values for x.... and then see if you can have some sort of simultaneous equation...

..im totally int he dark and wish i was still at home to dig up the maths books...its so weird cus i now see maths like a puzzle at the back of the paper with the sudoku and the crossword... its enjoyable!

[bryantm3]

You show us how you think it through first. :D
well, i first changed 2/3 and 1/3 powers on my paper to the cubed root of x squared, and the cubed root of -9x. then, i cubed everything, which, now that i think about it, is kind of retarded. anyway, i came up with some weird answers like 54i and 53.543i, and i could never figure out the problem. it doesn't really matter because at the final i had an average of 101.2 and it doesn't matter what grade i get on it. but i figure since i'm student of the semester i should definitely know how to do this.

[Jan]

How would you solve something like this?
x^4 + p * x^2 + q = 0

[patrick]

is the power on the 9*x or just the x? If so, just solve for the new equation when you substitute y = ( x^(1/3) ) and solve for y instead. Then cube root the y answer. I did this and got two values that work...

Let me know what you get...

[jOHN rODRIGUEZ]

Thank goodness patrick beat us to it. I, being the beligerent parent, bought a bottle of whiskey last night and couldn't think straight.

[bryantm3]

is the power on the 9*x or just the x? If so, just solve for the new equation when you substitute y = ( x^(1/3) ) and solve for y instead. Then cube root the y answer. I did this and got two values that work...

Let me know what you get...

There's no y, though— there's only one equation, and it's a binomial.

[Jan]

You introduce a new variable y = x^(1/3).

[bryantm3]

so that would leave the equation at y^2 -9y +8, correct?
so y would equal 1 and 8. 1^3 is 1, 8^3 is 512. thanks, that makes a lot more sense. i wonder why the teacher never taught that? maybe i wasn't there that day.

[bas_I_am]

There's no y, though— there's only one equation, and it's a binomial.
X to the power of 2/3, minus 9x to the power of 1/3, plus eight equals zero.


x^(2/3) - 9x^(1/3) + 8 = ( ³√x)² - 9(³√x) + 8

set y = ³√x

( ³√x)² - 9(³√x) + 8 →y² - 9y + 8 = 0 → ( y - 8 )( y - 1) = 0

r = { 8, 1 }

r 8 → ³√x → 8³ = x = 512 → ( ³√512)² - 9(³√512) + 8 = 8² - 9(8) + 8 = 64 - 72 + 8 = 0
r 1 → = ³√x → 1³ = x = 1 → ( ³√1)² - 9(³√1) + 8 = 1² - 9(1) + 8 = 1 - 9 + 8 = 0

[Dunwho]

eee gaz.... now i defo dont miss maths class.

[bas_I_am]

How would you solve something like this?
x^4 + p * x^2 + q = 0

x^4 + p * x^2 + q = → (x²)² + px² + q = 0

let y = x² → y² + py + q, and apply quadratic formula (http://en.wikipedia.org/wiki/Quadratic_equation)
with D = p² + 4q → y = (-p ± √D)/2 → x = √((-p ± √D)/2)

[sola sistim]

WHAT THE FUCK IS THIS?

and i thought i wasnt too bad at maths..
never actually done anything like that. didnt even know you could add letters.. hahah

[big screen satellite]

does Wolverine appear in x^4 + p * x^2 + q = → (x²)² + px² + q = 0?