Why is $frac{15sqrt[4]{125}}{sqrt[4]{5}}$ $15sqrt{5}$ and not $15sqrt[4]{25}$?












1














I have an expression I am to simplify:



$$frac{15sqrt[4]{125}}{sqrt[4]{5}}$$



I arrived at $15sqrt[4]{25}$. My textbook tells me that the answer is in fact $15sqrt{5}$. Here is my thought process:



$frac{15sqrt[4]{125}}{sqrt[4]{5}}$ = $frac{15*sqrt[4]{5}*sqrt[4]{25}}{sqrt[4]{5}}$ =



(cancel out $sqrt[4]{5}$ present in both numerator and denominator)
leaving:
$$15sqrt[4]{25}$$



Where did I go wrong and how can I arrive at $15sqrt{5}$?






algebra-precalculus







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  • 2




    The fourth root is the square root of the square root ...
    – Ethan Bolker
    2 days ago
















1














I have an expression I am to simplify:



$$frac{15sqrt[4]{125}}{sqrt[4]{5}}$$



I arrived at $15sqrt[4]{25}$. My textbook tells me that the answer is in fact $15sqrt{5}$. Here is my thought process:



$frac{15sqrt[4]{125}}{sqrt[4]{5}}$ = $frac{15*sqrt[4]{5}*sqrt[4]{25}}{sqrt[4]{5}}$ =



(cancel out $sqrt[4]{5}$ present in both numerator and denominator)
leaving:
$$15sqrt[4]{25}$$



Where did I go wrong and how can I arrive at $15sqrt{5}$?






algebra-precalculus







share|cite|improve this question


















  • 2




    The fourth root is the square root of the square root ...
    – Ethan Bolker
    2 days ago














1












1








1







I have an expression I am to simplify:



$$frac{15sqrt[4]{125}}{sqrt[4]{5}}$$



I arrived at $15sqrt[4]{25}$. My textbook tells me that the answer is in fact $15sqrt{5}$. Here is my thought process:



$frac{15sqrt[4]{125}}{sqrt[4]{5}}$ = $frac{15*sqrt[4]{5}*sqrt[4]{25}}{sqrt[4]{5}}$ =



(cancel out $sqrt[4]{5}$ present in both numerator and denominator)
leaving:
$$15sqrt[4]{25}$$



Where did I go wrong and how can I arrive at $15sqrt{5}$?






algebra-precalculus







share|cite|improve this question













I have an expression I am to simplify:



$$frac{15sqrt[4]{125}}{sqrt[4]{5}}$$



I arrived at $15sqrt[4]{25}$. My textbook tells me that the answer is in fact $15sqrt{5}$. Here is my thought process:



$frac{15sqrt[4]{125}}{sqrt[4]{5}}$ = $frac{15*sqrt[4]{5}*sqrt[4]{25}}{sqrt[4]{5}}$ =



(cancel out $sqrt[4]{5}$ present in both numerator and denominator)
leaving:
$$15sqrt[4]{25}$$



Where did I go wrong and how can I arrive at $15sqrt{5}$?






algebra-precalculus




algebra-precalculus






share|cite|improve this question













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asked 2 days ago









Doug Fir

2387




2387








  • 2




    The fourth root is the square root of the square root ...
    – Ethan Bolker
    2 days ago














  • 2




    The fourth root is the square root of the square root ...
    – Ethan Bolker
    2 days ago








2




2




The fourth root is the square root of the square root ...
– Ethan Bolker
2 days ago




The fourth root is the square root of the square root ...
– Ethan Bolker
2 days ago










3 Answers
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1














$$frac{sqrt[4] {125}}{sqrt[4] 5} = frac{sqrt[4] {5^3}}{sqrt[4] 5} = sqrt[4]{5^2} = 5^{frac{2}{4}} = 5^{frac{1}{2}} = sqrt 5$$



Even quicker, $sqrt[4]{25}$ means $sqrt{sqrt{25}}$, which becomes $sqrt{5}$.






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    2














    Write $$sqrt[4]{frac{125}{5}}=sqrt[4]{25}=sqrt{5}$$






    share|cite|improve this answer





























      2














      It turns out that $sqrt[4]{25}=sqrt{5}$. This is because $25=5^2$, so that $sqrt[4]{25}=sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=sqrt{5}.$ So, you are correct, as is the book.






      share|cite|improve this answer





















      • Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
        – Doug Fir
        2 days ago











      Your Answer





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      3 Answers
      3






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      3 Answers
      3






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      active

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      1














      $$frac{sqrt[4] {125}}{sqrt[4] 5} = frac{sqrt[4] {5^3}}{sqrt[4] 5} = sqrt[4]{5^2} = 5^{frac{2}{4}} = 5^{frac{1}{2}} = sqrt 5$$



      Even quicker, $sqrt[4]{25}$ means $sqrt{sqrt{25}}$, which becomes $sqrt{5}$.






      share|cite|improve this answer


























        1














        $$frac{sqrt[4] {125}}{sqrt[4] 5} = frac{sqrt[4] {5^3}}{sqrt[4] 5} = sqrt[4]{5^2} = 5^{frac{2}{4}} = 5^{frac{1}{2}} = sqrt 5$$



        Even quicker, $sqrt[4]{25}$ means $sqrt{sqrt{25}}$, which becomes $sqrt{5}$.






        share|cite|improve this answer
























          1












          1








          1






          $$frac{sqrt[4] {125}}{sqrt[4] 5} = frac{sqrt[4] {5^3}}{sqrt[4] 5} = sqrt[4]{5^2} = 5^{frac{2}{4}} = 5^{frac{1}{2}} = sqrt 5$$



          Even quicker, $sqrt[4]{25}$ means $sqrt{sqrt{25}}$, which becomes $sqrt{5}$.






          share|cite|improve this answer












          $$frac{sqrt[4] {125}}{sqrt[4] 5} = frac{sqrt[4] {5^3}}{sqrt[4] 5} = sqrt[4]{5^2} = 5^{frac{2}{4}} = 5^{frac{1}{2}} = sqrt 5$$



          Even quicker, $sqrt[4]{25}$ means $sqrt{sqrt{25}}$, which becomes $sqrt{5}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          KM101

          5,3811423




          5,3811423























              2














              Write $$sqrt[4]{frac{125}{5}}=sqrt[4]{25}=sqrt{5}$$






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                2














                Write $$sqrt[4]{frac{125}{5}}=sqrt[4]{25}=sqrt{5}$$






                share|cite|improve this answer
























                  2












                  2








                  2






                  Write $$sqrt[4]{frac{125}{5}}=sqrt[4]{25}=sqrt{5}$$






                  share|cite|improve this answer












                  Write $$sqrt[4]{frac{125}{5}}=sqrt[4]{25}=sqrt{5}$$







                  share|cite|improve this answer












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                  answered 2 days ago









                  Dr. Sonnhard Graubner

                  73.3k42865




                  73.3k42865























                      2














                      It turns out that $sqrt[4]{25}=sqrt{5}$. This is because $25=5^2$, so that $sqrt[4]{25}=sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=sqrt{5}.$ So, you are correct, as is the book.






                      share|cite|improve this answer





















                      • Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
                        – Doug Fir
                        2 days ago
















                      2














                      It turns out that $sqrt[4]{25}=sqrt{5}$. This is because $25=5^2$, so that $sqrt[4]{25}=sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=sqrt{5}.$ So, you are correct, as is the book.






                      share|cite|improve this answer





















                      • Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
                        – Doug Fir
                        2 days ago














                      2












                      2








                      2






                      It turns out that $sqrt[4]{25}=sqrt{5}$. This is because $25=5^2$, so that $sqrt[4]{25}=sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=sqrt{5}.$ So, you are correct, as is the book.






                      share|cite|improve this answer












                      It turns out that $sqrt[4]{25}=sqrt{5}$. This is because $25=5^2$, so that $sqrt[4]{25}=sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=sqrt{5}.$ So, you are correct, as is the book.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 2 days ago









                      Antonios-Alexandros Robotis

                      9,49241640




                      9,49241640












                      • Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
                        – Doug Fir
                        2 days ago


















                      • Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
                        – Doug Fir
                        2 days ago
















                      Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
                      – Doug Fir
                      2 days ago




                      Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
                      – Doug Fir
                      2 days ago


















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