simplifying a square- root radical

sammielou714 said:
i need help with simplifying a square- root radical of 3 and the square root of 8

Hi sammielou,

I don't quite understand what it is you need. '..simplyifing a square- root radical of 3' means \(\displaystyle \sqrt{3}\) and it cannot be simplified.

'...the square root of 8', on the other hand can be simplified this way:

\(\displaystyle \sqrt{8}=\sqrt{4 \cdot 2}=2\sqrt{2}\)

If you're asking for some combination of these two radicals, add, subtract, multiply, divide, then I suggest you use grouping symbols and be more specific in your explanation.
 
Maybe you are after the "cube root" of 8?
\(\displaystyle \sqrt[3]{8} =\sqrt[3]{2\cdot2\cdot2}\)
Can you finish?
 
easilyconfuzzeled said:
masters said:
sammielou714 said:
\(\displaystyle \sqrt{8}=\sqrt{4 \cdot 2}=2\sqrt{2}\)
im doing this stuff in my math class also and im condused on why you put a 2 on the outside of the radical?

Hi easilyconfuzzled,

Well, it breaks down this way. See if you can follow.

\(\displaystyle \sqrt{8}=\sqrt{4 \cdot 2}=\sqrt{4} \cdot \sqrt{2}=2 \cdot \sqrt{2}=2\sqrt{2}\)

You look for the largest perfect square factor of the radicand (the value under the radical).
In this case, it's 4.

You factor the radicand into this largest perfect square factor and whatever is left.
In this case it's \(\displaystyle 4 \cdot 2\)

Then, you take the square root of the perfect square factor and put the result in front of radical which now contains only the non-perfect square factor of the original radicand.

Here's another example.

\(\displaystyle \sqrt{48}=\sqrt{16 \cdot 3}=\sqrt{16} \cdot \sqrt{3}=4\sqrt{3}\)

Try some examples yourself. It's fun.
 
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