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ummmm okay, so the first thing your gonna do is subtract one from both sides

is that the equation?

the equation is x^(2/5)+x^(1/5)+1=3

how do i subtract 3 from the left side? do i plug in the numbers first?

x^(2/5) + x^(1/5) + 1 = 3
x^(2/5) + x^(1/5) - 2 = 0

oh wait, your right :D do your thing XD

so its 32 and 1 that equals 3 right?

32^2/5+1^1/5-2=0

hello

u^2+u=2
u(u+1)=2
u=1

how to i solve for x?
x^(1/5)=1?
is that how i do it?

if that is the value of u, then yes, there should be 2 values for u though

how do I get two values for u then?

they aren't using the same kind of numbers i am. what am i doing wrong?

Refer to the attachment from Mathematica 9.

It provides the right answer.

where is the answer in it?

none of your options are going to work because 1 is the only solution to this equation.

that's what i thought but it doesn't give me an option for just 1.

1?

i still don't understand any of this.
:(

i figured it out!!

ok. What did you get?

-32,1

yeah that's right :)