Question

Anyone good at Radicals? Solve these Please!

Answer 1

1. https://gcs490.brainhoney.com/Resource/14610120,45A,0,0/Assets/flvs/algebra2_v10_gs-xml/res0049/04_02_01.gif
2. https://gcs490.brainhoney.com/Resource/14610120,45A,0,0/Assets/flvs/algebra2_v10_gs-xml/res0049/04_02_16.gif
3. https://gcs490.brainhoney.com/Resource/14610120,45A,0,0/Assets/flvs/algebra2_v10_gs-xml/res0049/04_02_16.gif
4. https://gcs490.brainhoney.com/Resource/14610120,45A,0,0/Assets/flvs/algebra2_v10_gs-xml/res0049/04_04_06.gif
5. https://gcs490.brainhoney.com/Resource/14610120,45A,0,0/Assets/flvs/algebra2_v10_gs-xml/res0049/04_04_16.gif
6. https://gcs490.brainhoney.com/Resource/14610120,45A,0,0/Assets/flvs/algebra2_v10_gs-xml/res0049/04_07_01.gif
7. https://gcs490.brainhoney.com/Resource/14610120,45A,0,0/Assets/flvs/algebra2_v10_gs-xml/res0049/04_08_01.gif

Answer 2

Kaboooom :D

Answer 3

Too many. I'll show you how to do the first one, anyway.

\(\sqrt[3]{128}\) is the cube root of 128, that is, what number cubed is 128?

This isn't an integer, so you have to factor it and simplify it. You'll notice that
\[\sqrt[3]{128} = \sqrt[3]{64 \times 2} = \sqrt[3]{4^3 \times 2} = 4\sqrt[3]{2}\]

Answer 4

Errr, Confusing

Answer 5

What part is confusing?

Answer 6

All of it mainly haha

Answer 7

That doesn't really help me explain any of it. D:

Answer 8

Could you give me another example?

Answer 9

Ok, I'll do a simpler one like \(\sqrt{8}\)
We can factor 8 as \(2 \times 2 \times 2\).
This can also be written as \(2^2 \times 2\)
Thus, \(\sqrt{8} = \sqrt{2^2 \times 2}\)

It's a rule that \(\sqrt{ab} = \sqrt{a} \times \sqrt{b}\)
That means that \(\sqrt{2^2\times2} = \sqrt{2^2}\sqrt{2}\)
Simplifying that, it's \(2\sqrt{2}\)