Simplifying expressions with real and imaginary numbers

[andyleija]

The directions say "Simplify these expressions." I'm hoping posting pictures is alright. I didn't really see anything bad about it in the rules.

1.

Since there are two negative signs, I'm having a lot of trouble getting started here. I'm not sure where I'm supposed to put the i's and what to do with them.

2.

I assumed this is more of an issue of moving the variable out of the way. After moving the two g's and moving everything else under one cube root, what I've got is g^2xy(16y)^(1/3)

3.

After moving the g, I got g sqrt (-20r/5r) = g sqrt (-4r) = 2ig sqrt (r). I'm not sure if the i and g are in the right place and if I'm still supposed to have the sqrt (r).

4.

My biggest problem here is getting started. I've got no idea as to how to simplify this expression.

5.

I've done the problems two ways. If I'm not supposed to remove the g at all, I simplify the 10/8 and 2^-6/2^-2, which gives me 5/4g2^8. If I remove the g first, I simplify the 10/8 and 2^-6/2^-2, which gives me (5*2^8)/4 = (5*256)/4 = 1280/4 = 320.

6.

Here I added the exponents to give me t^(3 + n - 3) = t^n.

Thanks in advance for the help!

 

[Loren]

1. First steps... \(\displaystyle \sqrt{-1(4g+9)}=\sqrt{-1}\cdot\sqrt{4g+9}\) and you continue

2. They are all cube roots so to get a factor out from under the radical sign, it must be a perfect cube. For instance...\(\displaystyle \sqrt[3]{4x^4}\cdot \sqrt[3]{4x^4}=\sqrt[3]{16x^8}=\sqrt[3]{8x^6\cdot 2x^2}=2x^2\sqrt[3]{2x^2}\)

3. Do the r's cancel? Can you recuce 20/5? You have to be careful when working with negative signs under the radical. Don't cancel the negative values. It is best to rewrite your expression in i form first, then do the simplification. For instance...\(\displaystyle \sqrt{\frac{-2}{7}}\cdot \sqrt{\frac{21}{-4}}=i\sqrt{\frac{2}{7}}\cdot i\sqrt{\frac{21}{4}}=i^2\sqrt{\frac{3}{2}}=-\sqrt{\frac{3}{2}}\) not \(\displaystyle +\sqrt{\frac{3}{2}}\) which you would have gotten had you canceled the negatives under the radical sign.

4. \(\displaystyle \frac{x^a}{x^b}=x^{a-b}\)

5. You don't just "remove" things such as the g. This is a division problem. What is \(\displaystyle \frac{g}{2g}\)? I hope you don't get 2. Maybe 1 over 2 since g/g = 1?

6. Correct.