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.

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

3. Oops I meant simplified versions.

4. Is $\displaystyle 9x^2 + 2\sqrt{3y} - 3x^2\sqrt{3y}$ 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).

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

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

7. 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.

8. 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
$\displaystyle 9x^2 + 2 \sqrt{3y} - 3x^2 \sqrt{3y} = 9x^2 + (2 - 3x^2)\sqrt{3y}$

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

-Dan

9. Originally Posted by anthonye
The problem is $\displaystyle 4x\sqrt{12x^2y}+\sqrt{3x^4y}-x^2\sqrt{3y}$

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

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

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

-Dan

11. 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