My original question was poorly constructed and has led to some exasperation. I apologize, and here try to clarify my original dilemma.

Without LaTex, how do I express the Radical Sign?
My problem is: I have an open parenthesis... "3x"... then a Radical Sign with "4y" as a Radicand... close parenthesis... ^2

When I simplify this, the answer is: 36x^2y which I have a problem with for reasons following: I don't understand how the ^2 in the original term relates to the 36x because:
Take the following (from my book) for an example: (-2)^4 & -2^4 In the first term, (-2) is raised to the 4th power. In the second term, the rule says: "Without the parentheses, the exponent 4 applies only to the base 2, not to the minus sign."
Simplifying the original (above) I get: 3x^2 * 4y No Radical and no Parentheses. Without the parentheses, the exponent ^2 should apply only to the x. According to the rule, the 3 is not to be squared.... so how does the 36 get into the answer when the 3 is never squared and multiplied by 4?
Thank you. John

Originally Posted by John Whitaker
How do I express: an Index of ("3x"... a Radicand of "4y")^2 ?
I'm sorry, but I have no idea. What do you mean?

"Index" generally refers, in the context of radicals, to the type of root. For instance, a square root has an index of 2; a cube root has an index of 3. But I see no root indicated here. You refer to "radicand", but what is it inside?

"Index" can also refer (in British-English) to exponents. Since much of your post refers to confusion regarding how to deal with the square, this may be what you mean. But what do you mean by "expressing"? Were you given values for x and y, and you're supposed to "evaluate"?

Do you mean any of the following?

. . . . .$\L \left(\sqrt[3x]{4y}\right) ^2$

. . . . .$\L 3x \left(\sqrt{4y}\right) ^2$

. . . . .$\L \left( 3x \sqrt{4y}\right) ^2$

. . . . .$\L \left({\left( \sqrt{4y}\right)}^2 \right) ^{3x}$

Thank you.

Eliz.

3. Eliz,
#3 on your list is accurate. Thank you. John

$\sqrt {4y} = 2\sqrt y \quad \& \quad \sqrt {4y} \not= - 2\sqrt y .$

Please note that $\sqrt 4 = 2\quad \& \quad \sqrt 4 \not= \pm 2\quad \mbox{because} \quad \sqrt 4 > 0$

5. PKA,
Thank you.
The radical expression I refer to is the one Liz displays in her reply, third down on her list. The answer to that is: 36x^2y

6. Where are you getting a minus sign?
There is no minus in that problem whatsoever!
$\L \left( {3x\sqrt {4y} } \right)^2 = \left( {3x} \right)^2 \left( {\sqrt {4y} } \right)^2 = \left( {9x^2 } \right)\left( {4y} \right) = 36x^2 y$.

7. Originally Posted by John Whitaker
I do not have the program that allows all the math expressions to be displayed as you do...
Your browser isn't allowing you to input the LaTeX commands...?

Eliz.

8. PKA,
In your last LaTex reply, the first term after the first equal sign shows parentheses in the term: (3x)^2. The book I took this problem from DOES NOT have 3x in parentheses. This caused my confusion, and the reason I bring it to you, because my understanding of the rule is: without the parentheses, the exponent 2 would only apply to the x... not to the 3x.
My book does not display parentheses around the 3x as your LaTex version does. I have to assume the book is in error in omitting parentheses that you include. If your parentheses are appropriate, then I understand what's happening. Thanks.
John
(I will assume you are correct.)

9. If the third variant I'd posted was correct, and if the instructions were to "simplify the expression", then, yes, the answer would be 9x<sup>2</sup>4y = 36x<sup>2</sup>y.

Eliz.

10. Here are some more examples.$\L \begin{array}{l} \left( {2xy^2 z^3 } \right)^4 = \left( 2 \right)^4 \left( x \right)^4 \left( {y^2 } \right)^4 \left( {z^3 } \right)^4 = 16x^4 y^8 z^{12} \\ \left( { - xy^2 } \right)^3 = - x^3 y^6 \\ \left( { - xy^2 } \right)^4 \left( {x^3 y^2 } \right)^3 = \left( {x^4 y^8 } \right)\left( {x^9 y^6 } \right) = x^{13} y^{14} \\ \end{array}$

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