1. ## A few Radical Problems

Sorry guys ... just when I think I have the hang of these I have a few that jump out at me and make me start questioning what the heck I am doing.

My first problem is to mulitply and simplify assuming all variables in a radicand represent positive real numbers. I made the checks because I am not able to create the great square root symbols you can make.

√6x^3 * √4x^2

Second Problem is to add or subtract as indicated:I am not sure how to get this one to type correctly so I will say it looks like an exponent above and before the square root symbol. so I will type it above
3 3
√27x^4 + √xy^6

Third one I would think how many I have done of these I would have gotten this one correct. No laughing please!!

it says to express each number in terms of i and simplify if possible.
√-63

2. Originally Posted by kbryant05

√6x^3 * √4x^2
Note that,
$\displaystyle \sqrt{6x^3}=\sqrt{6x\cdot x^2}=\sqrt{6x}\cdot \sqrt{x^2}=x\sqrt{6x}$
Also note that,
$\displaystyle \sqrt{4x^2}=\sqrt{4}\cdot\sqrt{x^2}=2x$
Therefore,
$\displaystyle (2x)\cdot (x\sqrt{6x})=2x^2\sqrt{6x}$

Originally Posted by kbryant05
3 3
√27x^4 + √xy^6
Note that,
$\displaystyle \sqrt[3]{27x^4}=\sqrt[3]{27\cdot x^3\cdot x}=\sqrt[3]{27}\cdot \sqrt[3]{x^3}\sqrt[3]{x}=3x\sqrt[3]{x}$
Also that,
$\displaystyle \sqrt[3]{xy^6}=\sqrt[3]{y^6}\sqrt[3]{x}=y^2\sqrt[3]{x}$
$\displaystyle 3x\sqrt[3]{x}+y^2\sqrt[3]{x}=(3x+y^2)\sqrt[3]{x}$
Originally Posted by kbryant05[/tex
√-63
This is a trick I always told my classmates. Do the problem,
$\displaystyle \sqrt{63}$ then worry about the negative.
You have,
$\displaystyle \sqrt{63}=\sqrt{9\cdot 7}=\sqrt{9}\cdot \sqrt{7}$
Thus,
$\displaystyle 3\sqrt{7}$
Now introduce the "i" because the number was negative.
$\displaystyle 3i\sqrt{7}$

This is my 13th post!!!

3. ## Brilliant!!!

Thank you!!! I feel like an idiot just when I get so close I miss something. I started to do the FOIL method on the first one and went way beyond where I should have been going. Thanks for making it look so easy!!