how to simplify this radical fraction: 4th-root[ 225 / 81 ]

allegansveritatem

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This is not, strictly speaking, an algebra problem, but it was included in the problems given at the end of a section in my algebra book. I am being asked to simplify.Here is problem or expression or whatever it is:
problem.PNG

Here is book solution:
solution.PNG

Here is what I did:
mysolution.jpg

I know how to reduce this sort of expression when the exponent of something in the radicand is the same as the index or greater. But...how do you reduce something where the exponent is 2 and the index is 4? Now, I know that this is true because I checked it on a calculator. And I think I found that any time there is an exponent of 2 and an index of 4 you can subtract the lesser from the greater and result will be the new exponent of that element of the radicand. BUT: Try this with and exponent of 2 and an index of 5 and truth will forsake your results.
So, I am asking: How did the author get from the fourth root of 15^2 to the square root of 15?
 
This is not, strictly speaking, an algebra problem, but it was included in the problems given at the end of a section in my algebra book. I am being asked to simplify.Here is problem or expression or whatever it is:
View attachment 10649

Here is book solution:
View attachment 10650

Here is what I did:
View attachment 10651

I know how to reduce this sort of expression when the exponent of something in the radicand is the same as the index or greater. But...how do you reduce something where the exponent is 2 and the index is 4? Now, I know that this is true because I checked it on a calculator. And I think I found that any time there is an exponent of 2 and an index of 4 you can subtract the lesser from the greater and result will be the new exponent of that element of the radicand. BUT: Try this with and exponent of 2 and an index of 5 and truth will forsake your results.
So, I am asking: How did the author get from the fourth root of 15^2 to the square root of 15?
You are being asked 'what times itself 4 times gives 152'. Since sqrt(15)*sqrt(15)*sqrt(15)*sqrt(15)=152, the answer to my question and yours is sqrt(15).
 
This is not, strictly speaking, an algebra problem, but it was included in the problems given at the end of a section in my algebra book. I am being asked to simplify.Here is problem or expression or whatever it is:
View attachment 10649

Here is book solution:
View attachment 10650

Here is what I did:
View attachment 10651

I know how to reduce this sort of expression when the exponent of something in the radicand is the same as the index or greater. But...how do you reduce something where the exponent is 2 and the index is 4? Now, I know that this is true because I checked it on a calculator. And I think I found that any time there is an exponent of 2 and an index of 4 you can subtract the lesser from the greater and result will be the new exponent of that element of the radicand. BUT: Try this with and exponent of 2 and an index of 5 and truth will forsake your results.
So, I am asking: How did the author get from the fourth root of 15^2 to the square root of 15?
Taking the fourth root is the same as raising the radicand to the power 1/4.
So, \(\displaystyle \sqrt[4]{15^2} =(15^2)^{\frac{1}{4}}=15^{\frac{1}{2}} =\sqrt{15}\)

You don't subtract, you divide, for the reason shown above
 
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Taking the fourth root is the same as raising the radicand to the power 1/4.
So, \(\displaystyle \sqrt[4]{15^2} =(15^2)^{\frac{1}{4}}=15^{\frac{1}{2}} =\sqrt{15}\)

You don't subtract, you divide, for the reason shown above
This is a valid response but I feel that this should come after the true definition of \(\displaystyle \sqrt[4]{15^2}\)
 
You are being asked 'what times itself 4 times gives 152'. Since sqrt(15)*sqrt(15)*sqrt(15)*sqrt(15)=152, the answer to my question and yours is sqrt(15).
Right. I didn't think it through like this for some reason. Thanks
 
Taking the fourth root is the same as raising the radicand to the power 1/4.
So, \(\displaystyle \sqrt[4]{15^2} =(15^2)^{\frac{1}{4}}=15^{\frac{1}{2}} =\sqrt{15}\)

You don't subtract, you divide, for the reason shown above

This is something new to me. I haven't reached this in my text, but I have been trying it out with the help of a calculator and, sure enough, it works! I will remember it. Thanks.
 
Here:

radicals - 1.jpg
If you change the radical expression into expression with fractional exponent it will that way. Then, since 225 and 81 are perfect squares, we express them as 152 and 92 . Multiply their exponents (152(1/4) and 92(1/4)) and we would get 151/2 / 91/2. change them back to radicals and we will get (√15)/3. It turns out that 81 is a perfect 4th-root. Tell me which part confuses you.
 
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If I were doing this, I would factor \(\displaystyle 225~\&~81\).
\(\displaystyle 225=3^2\cdot 5^2~\&~81=3^4\)

\(\displaystyle \begin{align*}\sqrt[4]{\dfrac{225}{81}}&=\sqrt[4]{\dfrac{3^2\cdot 5^2}{3^4}}\\&=\dfrac{\sqrt[4]{3^2\cdot 5^2}}{3} \\&=\dfrac{\sqrt{15}}{3} \end{align*}\)
 
If I were doing this, I would factor \(\displaystyle 225~\&~81\).
\(\displaystyle 225=3^2\cdot 5^2~\&~81=3^4\)

\(\displaystyle \begin{align*}\sqrt[4]{\dfrac{225}{81}}&=\sqrt[4]{\dfrac{3^2\cdot 5^2}{3^4}}\\&=\dfrac{\sqrt[4]{3^2\cdot 5^2}}{3} \\&=\dfrac{\sqrt{15}}{3} \end{align*}\)

That's absolutely right for 81 but for 225, if I didn't know much about radicals, I would get confused if I factored 225 into 32 * 52 . is there any other way to make this easier than my calculation and yours?
 
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That's absolutely for 81 but for 225, if I didn't know much about radicals, I would get confused if I factored 225 into 32 * 52 . is there any other way to make this easier than my calculation and yours?
The derivation you gave in post 7 is just about as simple as this one is going to get.

-Dan
 
yeah. but what do you think about the 225 being factored to 3^2 * 5^2?
\(\displaystyle 3^2\cdot 5^2=9\cdot 25=225\)

You have problems with simple arithmetic of fractions.

\(\displaystyle \large \sqrt[4]{3^2\cdot 5^2}=\left(3^2\cdot 5^2\right)^{\frac{1}{4}}= 3^{\frac{1}{2}}\cdot 5^{\frac{1}{2}}=\sqrt{15}\)
 
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\(\displaystyle 3^2\cdot 5^2=9\cdot 25=225\)

You have problems with simple arithmetic of fractions.

\(\displaystyle \large \sqrt[4]{3^2\cdot 5^2}=\left(3^2\cdot 5^2\right)^{\frac{1}{4}}= 3^{\frac{1}{2}}\cdot 5^{\frac{1}{2}}=\sqrt{15}\)

sorry but the problem is you don't understand my question. anyway, I know that.
 
All questions from students are relevant unless they are rude/disrespectful like this jtaya0622
Jomo

This was triggered by response # 7, which was not a question but a purported answer to the original questions. I do not deem questions by a responder to be questions by a student.
 
Jomo

This was triggered by response # 7, which was not a question but a purported answer to the original questions. I do not deem questions by a responder to be questions by a student.
Does that matter? He has a question. He has a right to be helped. That's what we do here. His rudeness...That's another matter.

-Dan
 
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