Confused - Radical Expression

Nov 2009
21
0
California, United States
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:

\(\displaystyle
\sqrt{\frac{9x^7}{16y^8}}
\)

The solution which I thought would be right is:

\(\displaystyle
\frac{3x^6\sqrt{x}}{4y^7\sqrt{y}}
\)

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

\(\displaystyle
\frac{3x^2\sqrt{x}}{4y^4}
\)

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

MHF Helper
Jun 2008
16,216
6,764
North Texas
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:

\(\displaystyle
\sqrt{\frac{9x^7}{16y^8}}
\)

The solution which I thought would be right is:

\(\displaystyle
\frac{3x^6\sqrt{x}}{4y^7\sqrt{y}}
\)

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

\(\displaystyle
\frac{3x^2\sqrt{x}}{4y^4}
\)

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 ...



\(\displaystyle
\sqrt{\frac{9x^7}{16y^8}} = \sqrt{\frac{3x^3 \cdot 3x^3 \cdot x}{4y^4 \cdot 4y^4}} = \sqrt{\frac{(3x^3)^2 \cdot x}{(4y^4)^2}}
= \frac{\sqrt{(3x^3)^2} \cdot \sqrt{x}}{\sqrt{(4y^4)^2}} = \frac{3x^3 \sqrt{x}}{4y^4}\)
 
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Nov 2009
21
0
California, United States
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. :p Thank you much, I appreciate your help.

Colton
 
Nov 2009
21
0
California, United States
I'm sorry to bump, but I have another question. Here I have the following problem I've stumbled on:

\(\displaystyle
\frac{x^{2/5}y^{5/6}}{x^{-1/3}y^{1/2}}
\)

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

However, I got the answer of \(\displaystyle xy\sqrt[3]{xy}\)

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:

\(\displaystyle
\frac{xy^{9/6}}{xy^{1/6}}
\)

\(\displaystyle
xy^{8/6}
\)

\(\displaystyle
xy^{4/3}
\)

\(\displaystyle
\sqrt[3]{xy^4}
\)

\(\displaystyle
xy\sqrt[3]{xy}
\)

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

MHF Helper
Jun 2008
16,216
6,764
North Texas
I'm sorry to bump, but I have another question. Here I have the following problem I've stumbled on:

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

According to my workbook, the answer is \(\displaystyle x\sqrt[3]{y}\)
recheck the expression you typed ...

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

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


rule for division ...

\(\displaystyle \frac{x^a}{x^b} = x^{a-b}\)


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

\(\displaystyle x^{1} \cdot y^{\frac{1}{3}}\)

\(\displaystyle x \sqrt[3]{y}\)



in future, please start a new problem with a new thread.
 
Nov 2009
21
0
California, United States
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