Question

Find the real number solutions:
3x^4-27x^2+9x=x^3

Answer 1

3x^4-x^3-27^2+9x=0
-x^3(-3x+1)9x(-3x+1)=0
(-x^3+9x)(-3x+1)=0

Answer 2

-3x+1=0
x=1/3 . . .right?
And how can i solve for -x^3+9x=0?

Answer 3

\[3x^4-27x^2+9x=x^3\]\[3x^4-x^3-27x^2+9x=0\]\[x(3x^3-x^2-27x+9)=0\]is what I have so far, I'll continue but if you are right up to
-x^3+9x=0
-x(x^2-9)=0 --> x={-3,0,3}

Answer 4

Thats perfect, thank you so much! :D No need to continue. (:

Answer 5

cool, I was getting annoyed with this one LOL