anyone good at algebra?

[bas_I_am]

Originally Posted by bryantm3

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]

Originally Posted by Jan

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
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?