Question

Why are even and odd exponents significant when simplifying square root radical expressions?

Answer 1

What do you think?

\[\sqrt{x ^{6}}=\sqrt{x ^{3}\cdot x ^{3} }\]

\[\sqrt{x ^{7}}=\sqrt{x ^{3}\cdot x ^{3} \cdot{x}}\]

Do you see how it makes a difference in terms of simplifying the radical expression?

Answer 2

Ah, @pgpilot326 I see your point. That's a different angle on the question than how I was interpreting it, but a good point as well. :)

Answer 3

oops... didn't reallt read that question very well did i.

Answer 4

so what do I do

Answer 5

look at what @DebbieG wrote. what's the difference? what will happen with the first vs the second?

Answer 6

i'm going to delete what i first wrote because i don't want it to be a distraction... is that okay?

Answer 7

Remember, anything under the square root that is a square, gets to come be outside the square root. e.g. \[\sqrt{x ^{2}}=x\]\[\sqrt{12}=\sqrt{4\cdot 3 }=\sqrt{2^2\cdot 3 }\]
Since the 2 is squared, it gets to come out! The 3 is not a square, so it doesn't, it has to stay where it is:
\[\sqrt{12}=\sqrt{2^2\cdot 3 }=4\sqrt{3 }\]

Answer 8

oh ok ic now

Answer 9

Now how about if you have\[\sqrt{8}=\sqrt{4\cdot 2 }=\sqrt{2^2\cdot 2 }\]Again, part of what's under the square root sign (called a "radical" symbol, and what's under it is the "radicand") is a square, and part is not. The square comes out, the leftover does not.\[\sqrt{8}=\sqrt{4\cdot 2 }=\sqrt{2^2\cdot 2 }=2\sqrt{2 }\]