# Thread: How to get this other root?

1. ## How to get this other root?

Numbers are starting to look fuzzy again...

So, I have this practice problem:

$x^\frac{1}{2} - 3x^\frac{1}{3} = 3x^\frac{1}{6} - 9$

I was able to get the root x=27

$x^\frac{1}{3}(x^\frac{1}{6} - 3) = 3(x^\frac{1}{6} - 3)$
$x^\frac{1}{3} = 3$
$x = 27$

However, the book indicates that there are two roots: x = 27 and x = 729. I confirmed this with my calculator, but I am wondering how I was supposed to have figured out that other root. What did I miss here?

2. ## Re: How to get this other root?

Originally Posted by infraRed
$x^\frac{1}{2} - 3x^\frac{1}{3} = 3x^\frac{1}{6} - 9$
I was able to get the root x=27
$x^\frac{1}{3}(x^\frac{1}{6} - 3) = 3(x^\frac{1}{6} - 3)$
$x^\frac{1}{3} = 3$
$x = 27$
Let $y=x^{1/6}$
Can you solve $y^3-3y^2=3y-9~?$

3. ## Re: How to get this other root?

Another possibility is to rewrite:
$x^\frac{1}{3}(x^\frac{1}{6} - 3) = 3(x^\frac{1}{6} - 3)$
as
$x^\frac{1}{3}(x^\frac{1}{6} - 3) - 3(x^\frac{1}{6} - 3)=0$
$\Leftrightarrow (x^{\frac{1}{3}}-3)(x^{\frac{1}{6}}-3)=0$

Now use the fact that: $a\cdot b=0 \Leftrightarrow a=0 \ \mbox{or} \ b=0$
You can clearly see that there're two solutions.