radicals expression

**anthonye** · April 11th 2011, 03:07 AM

Hi All;
I have two solutions to the problem 9x^2 + 2sqrt3y - 3x^2sqrt3y they are

6x^2 + sqrt3y and 6x^2sqrt3y.

Please let me know which is right and how we got there.

Thanks.

**Prove It** · April 11th 2011, 03:09 AM

What you posted isn't even an equation, how can it have solutions?

**anthonye** · April 11th 2011, 03:16 AM

Oops I meant simplified versions.

**NOX Andrew** · April 11th 2011, 05:45 AM

Is $9x^2 + 2\sqrt{3y} - 3x^2\sqrt{3y}$ the expression in which you are interested? If so, then neither of the "simplifications" are equal to the original expression. In fact, the original expression can't be simplified much (if at all).

**anthonye** · April 11th 2011, 06:17 AM

The simplification on a web site is 6x^2sqrt3y.

**NOX Andrew** · April 11th 2011, 06:23 AM

What website? WolframAlpha doesn't offer any such expression as an equal expression.

**anthonye** · April 11th 2011, 06:32 AM

in browser search type (simplify radicals variables exponents) and scroll to the first pdf and its a question at the bottom at the bottom.
Problem 5.

**topsquark** · April 11th 2011, 07:35 AM

Originally Posted by anthonye

in browser search type (simplify radicals variables exponents) and scroll to the first pdf and its a question at the bottom at the bottom.
Problem 5.

Do you know how many search engines there are on the net? Give a link or something, will you?

However, no matter what the link is the only thing that can be done with this is
$9x^2 + 2 \sqrt{3y} - 3x^2 \sqrt{3y} = 9x^2 + (2 - 3x^2)\sqrt{3y}$

or
$9x^2 + 2 \sqrt{3y} - 3x^2 \sqrt{3y} = 3x^2(3 - \sqrt{3y} ) + 2 \sqrt{3y}$

-Dan

**anthonye** · April 11th 2011, 09:32 AM

http://http://www.germanna.edu/tutor..._exponents.pdf

heres the link.

**Plato** · April 11th 2011, 10:23 AM

Originally Posted by anthonye

http://http://www.germanna.edu/tutor..._exponents.pdf heres the link.

The problem is $4x\sqrt{12x^2y}+\sqrt{3x^4y}-x^2\sqrt{3y}$

I know how the got their answer. BUT it is wrong.
If you look at that expression we see that $y\ge 0$ but $x$ can be any real number.

Now why is it incorrect? Let $x=-3~\&~y=3$ the given expression evaluates to $270$ .
BUT $6x^2\sqrt{3y}=162$ . To be equal the values must be the same.
Here is the mistake. They said $4x\sqrt{12x^2y}=8x^2\sqrt{3y}$ . BUT IT DOES NOT.
This is what is correct: $4x\sqrt{12x^2y}=8x|x|\sqrt{3y}$ .

**topsquark** · April 11th 2011, 12:08 PM

Should I ask why the original post had nothing to do with this, or would it be smarter to leave it alone?

-Dan

**Plato** · April 11th 2011, 12:12 PM

Originally Posted by topsquark

Should I ask why the original post had nothing to do with this, or would it be smarter to leave it alone?-Dan

My guess is that the poster does not understand enough about radicals to post readable symbols.

**topsquark** · April 11th 2011, 12:31 PM

Originally Posted by Plato

My guess is that the poster does not understand enough about radicals to post readable symbols.

(snorts) That does it, I'm going back to my Bible study...

-Dan

**anthonye** · April 12th 2011, 02:18 AM

Ok thanks for the help everyone are there any sites that can be recommended that I can visit to learn/pratice more radicals?

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