## poopsiedoodle 3 years ago For problems 1–5, simplify the expression. Answers written in decimal form will not be accepted. Each of these problems is worth 1 point. "v/" is a radical by the way ___ 1.v/96 ______ 2. 8 v/63x^5 ___________ 3. v/128x^5y^2 ___ 4. ^3v/32 ________ 5. ^3v/56x^14

1. poopsiedoodle

So, I need them in radical form.

2. freewilly922

Factor everything first and then apply your radical rules... for example $\sqrt[3]{x^3} = x$

3. poopsiedoodle

Problem is that I have no idea what you're saying. ._.

4. freewilly922

Ok for example $\sqrt[8]{256x^4}$ becomes, if you factor $\sqrt[8]{2^{8}x^4}$ which then becomes $2\sqrt[8]{x^4}=2 x^{\frac{4}{8}}=2x^{1/2}=2\sqrt{2}$

5. freewilly922

This uses the idea that $\sqrt[a]{x} = x^{1/a}$

6. poopsiedoodle

just wondering, would you mind using bigger font? I can't see the exponents. To do that, put what you are saying in the curly braces in $$\huge\text{}\.) but take out the . at the end. 7. poopsiedoodle the ^ means exponents. like, 2^3 is 2 cubed. 8. poopsiedoodle \(\Large\text{But again, I have no idea how to do this.}$$

9. freewilly922

is ^3v/56x^14 supposed to be $\bigg(\sqrt{56x^{14}}\bigg)^3$ then?

10. poopsiedoodle

So, $$\huge\sqrt[8]{256x^4}$$ turns into $$\huge\sqrt[8]{2^{8}x^4}$$ which turns into $$\huge2\sqrt[8]{x^4}=2 x^{\frac{4}{8}}=2x^{1/2}=2\sqrt{2}$$ But how? $$\huge\text{:|}$$

11. freewilly922

Sorry that should be $2\sqrt{x}$

12. poopsiedoodle

and no, it's supposed to be $^{3}\sqrt{56x^{14}}$

13. freewilly922

ok the basic theorems that you need for these type of problems are that $\sqrt[a]{x} = x^{\frac{1}{a}}$ So for example $\large \sqrt[3]{x^{10}}\to x^{\frac{10}{3}}\to x^3x^{\frac{1}{3}}\to x^3\sqrt[3]{x}$

14. poopsiedoodle

I understand the first half of your example equation, but not the second half.

15. freewilly922

So any complicated radical you are given you can convert to exponents, use the rules of exponents shamelessly and then convert back to radical. So $\large \sqrt[3]{56x^{14}}\to \sqrt[3]{(7)(8)x^{14}}\to 7^{1/3}8^{1/3}x^{14/3}$ This then becomes $7^{1/3}8^{1/3}x^{14/3}\to 7^{1/3}(2^3)^{1/3}x^{12/3}x^{2/3}$ becomes $7^{1/3}2x^{4}x^{2/3}\to 2x^4 7^{1/3}x^{1/3}\to 2x^4\sqrt[3]{7x}$

16. poopsiedoodle

my brain is about to explode. e_o

17. freewilly922

So you don't understand how $x^{10/3} \to x^3x^{1/3}$? This results from the fact that $X^aX^b = X^{ab}$ and the reverse. So if you have $x^{10/3}$ you have $x^{9/3 + 1/3}\to x^{9/3}x^{1/3}\to x^3x^{1/3}$

18. poopsiedoodle

again, might I suggest the larger font size? Maybe Large would be a good replacement for huge though.