# I need help with an experssion. Not sure under what category it falls.

• Mar 4th 2010, 08:47 PM
Atreferre
I need help with an experssion. Not sure under what category it falls.
1
x^3

I need to find the square root of the above equation.

I found this in one of my homework questions and even after looking through all my notes I can't figure this one out. Can anyone perhaps get me started? I don't want to ask for the final answer because that wouldn't help me, but I'd very much appreciate a little bit of direction.
• Mar 4th 2010, 08:50 PM
Atreferre
Damn my lack of knowledge of these forums. The 1 is supposed to be above the x to the 3, as in a fraction.
• Mar 4th 2010, 09:24 PM
Bacterius
Don't worry, it is equivalent.

So, you want to find $\displaystyle \sqrt{\frac{1}{x^3}}$. You can apply the rules of square roots, and we get :

$\displaystyle \sqrt{\frac{1}{x^3}} = \frac{\sqrt{1}}{\sqrt{x^3}}$

Now, $\displaystyle \sqrt{1} = 1$, so ...

$\displaystyle \sqrt{\frac{1}{x^3}} = \frac{1}{\sqrt{x^3}}$

Note that $\displaystyle \sqrt{x^3} = (x^3)^{\frac{1}{2}}$. By applying the laws of exponents, we have $\displaystyle \sqrt{x^3} = (x^3)^{\frac{1}{2}} = x^{\frac{3}{2}}$. Therefore :

$\displaystyle \sqrt{\frac{1}{x^3}} = \frac{1}{\sqrt{x^3}} = \frac{1}{x^{\frac{3}{2}}}$

If you wish to have a fraction-free result, remember that $\displaystyle \frac{1}{a^b} = a^{-b}$. Thus :

$\displaystyle \sqrt{\frac{1}{x^3}} = \frac{1}{\sqrt{x^3}} = \frac{1}{x^{\frac{3}{2}}} = x^{\frac{-3}{2}}$.

Does this make sense ? I tried to put as much help as I could, if you still have questions don't hesitate to ask. This can be applied to any such simplification problem.