Confused - Radical Expression

**Noegddgeon** · May 15th 2010, 12:42 PM

Hello, everybody :]

I'm a little confused as to the solution of a radical expression. My goal is to simplify it as much as I can. Here is the expression:

$ \sqrt{\frac{9x^7}{16y^8}} $

The solution which I thought would be right is:

$ \frac{3x^6\sqrt{x}}{4y^7\sqrt{y}} $

Using Mathway.com, however, it's telling me that the answer is:

$ \frac{3x^2\sqrt{x}}{4y^4} $

I would greatly appreciate if someone could explain how I got the wrong answer so I may learn from my mistakes. :] Thank you all very much.

Colton

**skeeter** · May 15th 2010, 02:30 PM

Originally Posted by Noegddgeon

Hello, everybody :]

I'm a little confused as to the solution of a radical expression. My goal is to simplify it as much as I can. Here is the expression:

$ \sqrt{\frac{9x^7}{16y^8}} $

The solution which I thought would be right is:

$ \frac{3x^6\sqrt{x}}{4y^7\sqrt{y}} $

Using Mathway.com, however, it's telling me that the answer is:

$ \frac{3x^2\sqrt{x}}{4y^4} $

I would greatly appreciate if someone could explain how I got the wrong answer so I may learn from my mistakes. :] Thank you all very much.

Colton

recheck the "mathway.com" solution ...

$ \sqrt{\frac{9x^7}{16y^8}} = \sqrt{\frac{3x^3 \cdot 3x^3 \cdot x}{4y^4 \cdot 4y^4}} = \sqrt{\frac{(3x^3)^2 \cdot x}{(4y^4)^2}} = \frac{\sqrt{(3x^3)^2} \cdot \sqrt{x}}{\sqrt{(4y^4)^2}} = \frac{3x^3 \sqrt{x}}{4y^4}$

**Noegddgeon** · May 15th 2010, 02:51 PM

Skeeter,

Thank you for your reply. I must have entered the equation incorrectly into Mathway or something, because it gave me the one which I posted.... I find that rather strange.

Thank you much, I appreciate your help.

Colton

**Noegddgeon** · May 16th 2010, 02:33 PM

I'm sorry to bump, but I have another question. Here I have the following problem I've stumbled on:

$ \frac{x^{2/5}y^{5/6}}{x^{-1/3}y^{1/2}} $

According to my workbook, the answer is $x\sqrt[3]{y}$

However, I got the answer of $xy\sqrt[3]{xy}$

My basis for solving the problem was to add the exponents on the top and bottom by finding the least common factor, which led me to simplify it to this:

$ \frac{xy^{9/6}}{xy^{1/6}} $

$ xy^{8/6} $

$ xy^{4/3} $

$ \sqrt[3]{xy^4} $

$ xy\sqrt[3]{xy} $

I am pretty much a beginner at math still, so I may have made a beginner's mistake when solving this problem. I would greatly appreciate some assistance, and I hope my step-by-step posting helps show where I may have made that mistake. :] Thank you very much in advance.

Colton

**skeeter** · May 16th 2010, 02:53 PM

Originally Posted by Noegddgeon

I'm sorry to bump, but I have another question. Here I have the following problem I've stumbled on:

$ \frac{x^{2/\textcolor{red}{5}}y^{5/6}}{x^{-1/3}y^{1/2}} $

According to my workbook, the answer is $x\sqrt[3]{y}$

recheck the expression you typed ...

you sure it's not $\frac{x^{\frac{2}{\textcolor{red}{3}}} y^{\frac{5}{6}}}{x^{-\frac{1}{3}} y^{\frac{1}{2}}}$ ???

if this is the original expression, then I agree w/ the workbook's solution.

rule for division ...

$\frac{x^a}{x^b} = x^{a-b}$

$x^{\frac{2}{3} - (-\frac{1}{3})} \cdot y^{\frac{5}{6} - \frac{1}{2}}$

$x^{1} \cdot y^{\frac{1}{3}}$

$x \sqrt[3]{y}$

in future, please start a new problem with a new thread.

**Noegddgeon** · May 16th 2010, 03:08 PM

Skeeter,

Yes, that was a typo. My apologies. I greatly appreciate you showing me that problem, however, and I see my mistake now :] In the future, I will remember to start a new thread.

Colton

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