OpenStudy (anonymous):

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

4 years ago
OpenStudy (debbieg):

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?

4 years ago
OpenStudy (debbieg):

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. :)

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

so what do I do

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (debbieg):

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: $\sqrt{12}=\sqrt{2^2\cdot 3 }=4\sqrt{3 }$

4 years ago
OpenStudy (anonymous):

oh ok ic now

4 years ago
OpenStudy (debbieg):

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.$\sqrt{8}=\sqrt{4\cdot 2 }=\sqrt{2^2\cdot 2 }=2\sqrt{2 }$

4 years ago
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