# Unable to solve roots expression

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Jun 12th 2009, 11:28 AM
allyourbass2212
Unable to solve roots expression
Expression 1:$\displaystyle \frac{1}{\sqrt[5]{8xy^2}}$

My Work:
$\displaystyle \frac{1}{\sqrt[5]{8xy^2}}=\frac{1}{\sqrt[5]{8xy^2}}*\frac{\sqrt[5]{8^4x^4y^3}}{\sqrt[5]{ 8^4x^4y^3}}=\frac{4\sqrt4}{8xy}$

The only answer choices I have for this self test are
Answer: $\displaystyle (8xy^2)^{-1/5}$

Expression 2:
$\displaystyle \frac{y^3}{\sqrt[4]y}$

The author states the correct answer for expression 2 is $\displaystyle x^{11/4}$, and yes they say "x" not "y"....either way not sure how to arrive at this answer either.

This is a self teaching algebra book, however the author decided to not show her work in any of the chapter reviews which is frustrating.

Thanks again!
• Jun 12th 2009, 12:00 PM
Amer
Quote:

Originally Posted by allyourbass2212
Expression 1:$\displaystyle \frac{1}{\sqrt[5]8xy^2}$

My Work:
$\displaystyle \frac{1}{\sqrt[5]8xy^2}$$\displaystyle =\frac{\sqrt[5]8^4x^4y^3}{\sqrt[5]8^4x^4y^3}=\frac{4\sqrt[5]4x^4y^3}{8xy}=\frac{\sqrt4x^4y^3}{2} The only answer choices I have for this self test are Answer: \displaystyle (8xy^2)^{-1/5} Expression 2: \displaystyle \frac{y^3}{\sqrt[4]y} The author states the correct answer for expression 2 is \displaystyle x^{11/4}, and yes they say "x" not "y"....either way not sure how to arrive at this answer either. This is a self teaching algebra book, however the author decided to not show her work in any of the chapter reviews which is frustrating. Thanks again! for the first one \displaystyle \frac{1}{\sqrt[5]{8xy^2}}=\left((8xy^2)\right)^{\frac{-1}{5}} since \displaystyle \frac{1}{A}=A^{-1} and \displaystyle \sqrt[n]{A}=A^{\frac{1}{n}} \displaystyle \frac{y^3}{\sqrt[4]{y}} =$$\displaystyle y^3 (\sqrt[4]{y})^{-1}$

$\displaystyle y^3 (\sqrt[4]{y})^{-1}= y^3(y^{\frac{-1}{4}})$

$\displaystyle y^{\frac{12}{4}+\frac{-1}{4}}=y^{\frac{11}{4}}$
• Jun 12th 2009, 02:08 PM
allyourbass2212
For
$\displaystyle \frac{y^3}{\sqrt[4]{y}}=$

Since this is a root thats cubed why can we not eliminate the nth root in the denominator, and write the denominator as the nth root of some quantity of the nth power.

e.g.

$\displaystyle \frac{y^3}{\sqrt[4]y} = \frac{y^3}{\sqrt[4]y}*\frac{\sqrt[4]y^3}{\sqrt[4]y^3}$
• Jun 12th 2009, 02:33 PM
yeongil
Quote:

Originally Posted by allyourbass2212
For
$\displaystyle \frac{y^3}{\sqrt[4]{y}}=$

Since this is a root thats cubed why can we not eliminate the nth root in the denominator, and write the denominator as the nth root of some quantity of the nth power.

e.g.

$\displaystyle \frac{y^3}{\sqrt[4]y} = \frac{y^3}{\sqrt[4]y}*\frac{\sqrt[4]y^3}{\sqrt[4]y^3}$

You could do that, but why? It would just be more work. You have two powers where the bases are the same. Using the properties of exponents, just subtract the exponents:

$\displaystyle \frac{y^3}{\sqrt[4]y} = \frac{y^3}{y^{1/4}} = y^{3 - 1/4} = y^{11/4}$

01
• Jun 12th 2009, 03:11 PM
allyourbass2212
Removed, due to incorrect expression see original post edited.
• Jun 12th 2009, 03:24 PM
yeongil
You state that this is the problem:
Quote:

Originally Posted by allyourbass2212
Expression 1:
$\displaystyle \frac{1}{\sqrt[5]8xy^2}$

and that this is the answer:
Quote:

The only answer choices I have for this self test are
Answer: $\displaystyle (8xy^2)^{-1/5}$
Something is wrong, because the problem, as you wrote it, doesn't simplify to that answer. Part of the confusion is whether the problem is supposed to be this:
$\displaystyle \frac{1}{\sqrt[5]8xy^2}\;\;\;{\color{red}Eq.1a}$
or this:
$\displaystyle \frac{1}{\sqrt[5]{8xy^2}}\;\;\;{\color{red}Eq.1b}$

See the difference? Amer assumes you meant Eq.1b, and it does simplify to the answer you wrote above.

You will need to double-check to see if you copied down the problem and answer correctly before we can proceed. If the problem is supposed to be Eq.1a, then yes, you would simplify by rationalizing the denominator. But if the problem is supposed to be Eq.1b, then you just need to rewrite using negative & fractional exponents.

01
• Jun 12th 2009, 03:32 PM
allyourbass2212
I meant 1.b, thanks for the clarification. But I still do not see why I cannot proceed as I did rationalizing the denominator.
• Jun 12th 2009, 03:56 PM
yeongil
Quote:

Originally Posted by allyourbass2212
I meant 1.b, thanks for the clarification. But I still do not see why I cannot proceed as I did rationalizing the denominator.

If we pretend that the problem was Eq.1a:
$\displaystyle \frac{1}{\sqrt[5]8xy^2}$
When rationalizing the denominator, you only deal with the 5th root of 5. You don't need to touch the x or y^2 because they are not underneath the radical sign:
\displaystyle \begin{aligned} \frac{1}{\sqrt[5]8xy^2} &= \frac{1}{\sqrt[5]8xy^2} \times \frac{\sqrt[5]{8^4}}{\sqrt[5]{8^4}} \\ &= \frac{\sqrt[5]{8^4}}{8xy^2} \\ &= \frac{4\sqrt[5]{4}}{8xy^2} \\ &= \frac{\sqrt[5]{4}}{2xy^2} \end{aligned}

But the problem is actually Eq.1b:
$\displaystyle \frac{1}{\sqrt[5]{8xy^2}}$

If you really want to rationalize the denominator, it would go like this:
\displaystyle \begin{aligned} \frac{1}{\sqrt[5]{8xy^2}} &= \frac{1}{\sqrt[5]{8xy^2}} \times \frac{\sqrt[5]{(8xy^2)^4}}{\sqrt[5]{(8xy^2)^4}} \\ &= \frac{\sqrt[5]{4096x^4 y^8}}{8xy^2} \\ &= \frac{4y\sqrt[5]{4x^4 y^3}}{8xy^2} \\ &= \frac{\sqrt[5]{4x^4 y^3}}{2xy} \end{aligned}

See how messy it is, compared to the answer given? (Wink)

$\displaystyle \frac{1}{\sqrt[5]{8xy^2}} = (8xy^2)^{-1/5}$

01
• Jun 12th 2009, 04:36 PM
allyourbass2212
Also, I do not understand your solutions yeongil & Amer

http://www.mathhelpforum.com/math-he...e3af1f69-1.gif

I do not understand how the y ^2 does not become part of the fraction, or where the -1 numerator comes from.

e.g.http://www.mathhelpforum.com/math-he...7cfed140-1.gif
in the case of this expression it seems to me the n = 5 and the m = 2.

thanks again
• Jun 12th 2009, 06:04 PM
Amer
Quote:

Originally Posted by allyourbass2212
Also, I do not understand your solutions yeongil & Amer

http://www.mathhelpforum.com/math-he...e3af1f69-1.gif

I do not understand how the y ^2 does not become part of the fraction, or where the -1 numerator comes from.

e.g.http://www.mathhelpforum.com/math-he...7cfed140-1.gif
in the case of this expression it seems to me the n = 5 and the m = 2.

thanks again

it comes from this
for any real numbers A you can write this

$\displaystyle \frac{1}{A}=A^{-1}$ so

$\displaystyle \frac{1}{x^2}=x^{-2}$

$\displaystyle \frac{y^3}{x^9}=\frac{x^{-9}}{y^{-3}}$

and

$\displaystyle \sqrt[n]{y}=y^{\frac{1}{n}}$

$\displaystyle \sqrt[n]{x^4}=x^{\frac{4}{n}}$ in general

$\displaystyle \sqrt[n]{x^m}=x^{\frac{m}{n}}$

mixed example

$\displaystyle \frac{1}{\sqrt[9]{x^4}}=x^{\frac{-4}{9}}$

$\displaystyle \frac{x^6}{\sqrt[10]{x^4}}=(x^6)(x^{\frac{-4}{10}})$

$\displaystyle \Rightarrow (x^{\frac{60}{10}})(x^{\frac{-4}{10}})$

$\displaystyle \Rightarrow x^{\frac{60-4}{10}}$

$\displaystyle \Rightarrow x^{\frac{56}{10}}$

$\displaystyle \Rightarrow x^{\frac{50}{10}+\frac{6}{10}}$

$\displaystyle \Rightarrow x^{5} (x^{\frac{6}{10}})=x^5(\sqrt[10]{x^6})$
• Jun 12th 2009, 06:28 PM
allyourbass2212
http://www.mathhelpforum.com/math-he...e3af1f69-1.gif

But why is the numerator 1 and not the two? e.g. 8xy^-2/5
• Jun 12th 2009, 06:36 PM
Amer
Quote:

Originally Posted by allyourbass2212
http://www.mathhelpforum.com/math-he...e3af1f69-1.gif

But why is the numerator 1 and not the two? e.g. 8xy^-2/5

I see

the power 2 is for y not for x

$\displaystyle \frac{1}{\sqrt[n]{x^4y^2z^7}}=(x^4y^2z^7)^{\frac{-1}{n}}$

but if
$\displaystyle \frac{1}{\sqrt[4]{x^3y^3}}=(xy)^{\frac{3}{4}}$ because the power of x and y is the same
• Jun 13th 2009, 05:59 AM
allyourbass2212
Quote:

Originally Posted by Amer
I see

the power 2 is for y not for x

$\displaystyle \frac{1}{\sqrt[n]{x^4y^2z^7}}=(x^4y^2z^7)^{\frac{-1}{n}}$

but if
$\displaystyle \frac{1}{\sqrt[4]{x^3y^3}}=(xy)^{\frac{3}{4}}$ because the power of x and y is the same

I just do not clearly see how these examples coincide with $\displaystyle \sqrt[n]{a^m}$

For instance wouldnt the $\displaystyle m$ in this case seems like it should be the $\displaystyle x^4$, instead its 1 and im not sure how you choose that.
$\displaystyle \frac{1}{\sqrt[n]{x^4y^2z^7}}=(x^4y^2z^7)^{\frac{-1}{n}}$

And in the other example you use the 3 as m. I assume because every exponent under the root is a 3.

$\displaystyle \frac{1}{\sqrt[4]{x^3y^3}}=(xy)^{\frac{3}{4}}$
• Jun 13th 2009, 08:54 AM
Amer
Quote:

Originally Posted by allyourbass2212
$\displaystyle \frac{1}{\sqrt[n]{x^4y^2z^7}}=(x^4y^2z^7)^{\frac{-1}{n}}$

And in the other example you use the 3 as m. I assume because every exponent under the root is a 3.

$\displaystyle \frac{1}{\sqrt[4]{x^3y^3}}=(xy)^{\frac{3}{4}}$

anyway I think with my example I made it more harder

I mistake in the second example

$\displaystyle \frac{1}{\sqrt[4]{x^3y^3}}= (xy)^{\frac{-3}{4}}$

Quote:

Originally Posted by allyourbass2212
I just do not clearly see how these examples coincide with $\displaystyle \sqrt[n]{a^m}$

For instance wouldnt the $\displaystyle m$ in this case seems like it should be the $\displaystyle x^4$, instead its 1 and im not sure how you choose that.

ok lats solve it in other way

$\displaystyle \frac{1}{\sqrt[5]{8xy^2}}$

$\displaystyle \frac{1}{(8xy^2)^{\frac{1}{5}}}$ since any n root you can write it like this
$\displaystyle \sqrt[n]{a} = a^{\frac{1}{n}}$

if it was in the denominator or if it was in the numentor like this

$\displaystyle \frac{1}{\sqrt[n]{a}}=\frac{1}{a^{\frac{1}{n}}}$
ok

$\displaystyle \frac{1}{(8xy^2)^{\frac{1}{5}}}$
now the improtant step

$\displaystyle \frac{1}{(8xy^2)^{\frac{1}{5}}} = (8xy^2)^{\frac{-1}{5}}$

since any fraction $\displaystyle \frac{1}{a} = a^{-1}$ for any a

a can be

$\displaystyle \frac{1}{(allyourbass2212)} = (allyourbass2212)^{-1}$ a can be anything ploynomial or function anything ..

$\displaystyle \frac{1}{\left((\sqrt[15]{t^8})log(sinx^2)cosy\right)}=\left((\sqrt[15]{t^8})log(sinx^2)cosy\right)^{-1}$

then if you can simplify it is ok like this

$\displaystyle \frac{x}{\sqrt{x}} = x(\frac{1}{x^{\frac{1}{2}}}) = x(x^{\frac{-1}{2}}) = x^{1+\frac{-1}{2}}=x^{\frac{1}{2}}$

it is clear or not ?
• Jun 13th 2009, 11:17 AM
allyourbass2212
Unfortunately I am still confused
$\displaystyle \sqrt[n]{a^m}$

http://www.mathhelpforum.com/math-he...48a91f42-1.gif = http://www.mathhelpforum.com/math-he...3536a87d-1.gif

I still do not understand why in the above expression $\displaystyle m = 1$ and more importantly why $\displaystyle m \neq 2$

An example like this makes sense $\displaystyle \sqrt[5]{x^3}=x^{3/5}$, but in the above expression it does not.
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last