# Square root with variables and exponents?

• Aug 27th 2009, 11:13 AM
memento mori
Square root with variables and exponents?
I honestly just drew a blank on this problem; it'd be great if you guys could walk me through how to solve it.

$\displaystyle \sqrt{a^5b^1}{^2}$
• Aug 27th 2009, 11:27 AM
RobLikesBrunch
Quote:

Originally Posted by memento mori
I honestly just drew a blank on this problem; it'd be great if you guys could walk me through how to solve it.

$\displaystyle \sqrt{a^5b^1}{^2}$

A square root can be written as an exponent: $\displaystyle \sqrt{a}=a^\frac{1}{2}$. I'm unsure of the exact reasoning, but you can sort of see why this works intuitively, as we know that $\displaystyle \sqrt{a}\cdot\sqrt{a} = a = a^\frac{1}{2} \cdot a^\frac{1}{2}$ (because we know that when multiply two of the same numbers with exponents, we add the exponents).

See if you can figure it out from here. If you can't, just tell me and I'll walk you through it.
• Aug 27th 2009, 11:32 AM
masters
Quote:

Originally Posted by memento mori
I honestly just drew a blank on this problem; it'd be great if you guys could walk me through how to solve it.

$\displaystyle \sqrt{a^5b^1}{^2}$

Hi momento mori,

$\displaystyle \sqrt{a^5b^{12}}=a^{\frac{5}{2}}b^{\frac{12}{2}}=a ^{\frac{4}{2}}\cdot a^{\frac{1}{2}}\cdot b^6=a^2b^6\sqrt{a}$
• Aug 27th 2009, 11:49 AM
memento mori
Thank you both very much; through reading both of your posts I was able to get the answer while thoroughly understanding the steps taken (and why they are taken) as well. :)