How can I simplify $sqrt{50y^8}$ to $5y^4sqrt{2}$?












0












$begingroup$


Question is in the title. My textbook has given me the question and solution but nothing in between. For this question I do not even know where to start.



In this chapter I've been learning about radicals, including the product and quotient rule, along with how to add and subtract radicals, rationalizing denominators and rational exponents.



Yet, for this particular question I'm struggling to even get started and understand what my first step would be?



I tried expressing the base as a exponent as opposed to a radical $50y^{2/8}$ to see if that would lead me down a path but it did not.



I find this question tricky because I am not sure which of the things I've been reading about in this chapter to apply here with a goal of simplifying the expression to $5y^4sqrt{2}$?






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I suppose you mean $sqrt{50y^8}$ in the title?
    $endgroup$
    – Yanko
    Jan 4 at 16:36










  • $begingroup$
    It looks like you are missing a square in the second expression. $(5y^4sqrt{2})^2$ gives you $25y^8cdot 2$
    $endgroup$
    – RScrlli
    Jan 4 at 16:37






  • 2




    $begingroup$
    Anyway, it looks like all you're missing the fact that for any two real numbers $a,b$ one has $sqrt{ab} = sqrt{a}sqrt{b}$. So for instance since $50=25*2$ we have $sqrt{50} = sqrt{25}sqrt{2} = 5cdot sqrt{2}$.
    $endgroup$
    – Yanko
    Jan 4 at 16:39










  • $begingroup$
    Thanks for catching that, yes sorry and I've updated the title now too.
    $endgroup$
    – Doug Fir
    Jan 4 at 16:39










  • $begingroup$
    Check this out math.stackexchange.com/questions/2047349/…
    $endgroup$
    – Yanko
    Jan 4 at 16:40
















0












$begingroup$


Question is in the title. My textbook has given me the question and solution but nothing in between. For this question I do not even know where to start.



In this chapter I've been learning about radicals, including the product and quotient rule, along with how to add and subtract radicals, rationalizing denominators and rational exponents.



Yet, for this particular question I'm struggling to even get started and understand what my first step would be?



I tried expressing the base as a exponent as opposed to a radical $50y^{2/8}$ to see if that would lead me down a path but it did not.



I find this question tricky because I am not sure which of the things I've been reading about in this chapter to apply here with a goal of simplifying the expression to $5y^4sqrt{2}$?






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I suppose you mean $sqrt{50y^8}$ in the title?
    $endgroup$
    – Yanko
    Jan 4 at 16:36










  • $begingroup$
    It looks like you are missing a square in the second expression. $(5y^4sqrt{2})^2$ gives you $25y^8cdot 2$
    $endgroup$
    – RScrlli
    Jan 4 at 16:37






  • 2




    $begingroup$
    Anyway, it looks like all you're missing the fact that for any two real numbers $a,b$ one has $sqrt{ab} = sqrt{a}sqrt{b}$. So for instance since $50=25*2$ we have $sqrt{50} = sqrt{25}sqrt{2} = 5cdot sqrt{2}$.
    $endgroup$
    – Yanko
    Jan 4 at 16:39










  • $begingroup$
    Thanks for catching that, yes sorry and I've updated the title now too.
    $endgroup$
    – Doug Fir
    Jan 4 at 16:39










  • $begingroup$
    Check this out math.stackexchange.com/questions/2047349/…
    $endgroup$
    – Yanko
    Jan 4 at 16:40














0












0








0





$begingroup$


Question is in the title. My textbook has given me the question and solution but nothing in between. For this question I do not even know where to start.



In this chapter I've been learning about radicals, including the product and quotient rule, along with how to add and subtract radicals, rationalizing denominators and rational exponents.



Yet, for this particular question I'm struggling to even get started and understand what my first step would be?



I tried expressing the base as a exponent as opposed to a radical $50y^{2/8}$ to see if that would lead me down a path but it did not.



I find this question tricky because I am not sure which of the things I've been reading about in this chapter to apply here with a goal of simplifying the expression to $5y^4sqrt{2}$?






algebra-precalculus







share|cite|improve this question











$endgroup$




Question is in the title. My textbook has given me the question and solution but nothing in between. For this question I do not even know where to start.



In this chapter I've been learning about radicals, including the product and quotient rule, along with how to add and subtract radicals, rationalizing denominators and rational exponents.



Yet, for this particular question I'm struggling to even get started and understand what my first step would be?



I tried expressing the base as a exponent as opposed to a radical $50y^{2/8}$ to see if that would lead me down a path but it did not.



I find this question tricky because I am not sure which of the things I've been reading about in this chapter to apply here with a goal of simplifying the expression to $5y^4sqrt{2}$?






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 16:39







Doug Fir

















asked Jan 4 at 16:34









Doug FirDoug Fir

4258




4258








  • 1




    $begingroup$
    I suppose you mean $sqrt{50y^8}$ in the title?
    $endgroup$
    – Yanko
    Jan 4 at 16:36










  • $begingroup$
    It looks like you are missing a square in the second expression. $(5y^4sqrt{2})^2$ gives you $25y^8cdot 2$
    $endgroup$
    – RScrlli
    Jan 4 at 16:37






  • 2




    $begingroup$
    Anyway, it looks like all you're missing the fact that for any two real numbers $a,b$ one has $sqrt{ab} = sqrt{a}sqrt{b}$. So for instance since $50=25*2$ we have $sqrt{50} = sqrt{25}sqrt{2} = 5cdot sqrt{2}$.
    $endgroup$
    – Yanko
    Jan 4 at 16:39










  • $begingroup$
    Thanks for catching that, yes sorry and I've updated the title now too.
    $endgroup$
    – Doug Fir
    Jan 4 at 16:39










  • $begingroup$
    Check this out math.stackexchange.com/questions/2047349/…
    $endgroup$
    – Yanko
    Jan 4 at 16:40














  • 1




    $begingroup$
    I suppose you mean $sqrt{50y^8}$ in the title?
    $endgroup$
    – Yanko
    Jan 4 at 16:36










  • $begingroup$
    It looks like you are missing a square in the second expression. $(5y^4sqrt{2})^2$ gives you $25y^8cdot 2$
    $endgroup$
    – RScrlli
    Jan 4 at 16:37






  • 2




    $begingroup$
    Anyway, it looks like all you're missing the fact that for any two real numbers $a,b$ one has $sqrt{ab} = sqrt{a}sqrt{b}$. So for instance since $50=25*2$ we have $sqrt{50} = sqrt{25}sqrt{2} = 5cdot sqrt{2}$.
    $endgroup$
    – Yanko
    Jan 4 at 16:39










  • $begingroup$
    Thanks for catching that, yes sorry and I've updated the title now too.
    $endgroup$
    – Doug Fir
    Jan 4 at 16:39










  • $begingroup$
    Check this out math.stackexchange.com/questions/2047349/…
    $endgroup$
    – Yanko
    Jan 4 at 16:40








1




1




$begingroup$
I suppose you mean $sqrt{50y^8}$ in the title?
$endgroup$
– Yanko
Jan 4 at 16:36




$begingroup$
I suppose you mean $sqrt{50y^8}$ in the title?
$endgroup$
– Yanko
Jan 4 at 16:36












$begingroup$
It looks like you are missing a square in the second expression. $(5y^4sqrt{2})^2$ gives you $25y^8cdot 2$
$endgroup$
– RScrlli
Jan 4 at 16:37




$begingroup$
It looks like you are missing a square in the second expression. $(5y^4sqrt{2})^2$ gives you $25y^8cdot 2$
$endgroup$
– RScrlli
Jan 4 at 16:37




2




2




$begingroup$
Anyway, it looks like all you're missing the fact that for any two real numbers $a,b$ one has $sqrt{ab} = sqrt{a}sqrt{b}$. So for instance since $50=25*2$ we have $sqrt{50} = sqrt{25}sqrt{2} = 5cdot sqrt{2}$.
$endgroup$
– Yanko
Jan 4 at 16:39




$begingroup$
Anyway, it looks like all you're missing the fact that for any two real numbers $a,b$ one has $sqrt{ab} = sqrt{a}sqrt{b}$. So for instance since $50=25*2$ we have $sqrt{50} = sqrt{25}sqrt{2} = 5cdot sqrt{2}$.
$endgroup$
– Yanko
Jan 4 at 16:39












$begingroup$
Thanks for catching that, yes sorry and I've updated the title now too.
$endgroup$
– Doug Fir
Jan 4 at 16:39




$begingroup$
Thanks for catching that, yes sorry and I've updated the title now too.
$endgroup$
– Doug Fir
Jan 4 at 16:39












$begingroup$
Check this out math.stackexchange.com/questions/2047349/…
$endgroup$
– Yanko
Jan 4 at 16:40




$begingroup$
Check this out math.stackexchange.com/questions/2047349/…
$endgroup$
– Yanko
Jan 4 at 16:40










1 Answer
1






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oldest

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2












$begingroup$

Use that:



$$sqrt{ab}=sqrt asqrt b$$
(when $a,b>0$)



Particularly useful here is:



$$sqrt{a^2b}=asqrt b$$






share|cite|improve this answer









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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Use that:



    $$sqrt{ab}=sqrt asqrt b$$
    (when $a,b>0$)



    Particularly useful here is:



    $$sqrt{a^2b}=asqrt b$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Use that:



      $$sqrt{ab}=sqrt asqrt b$$
      (when $a,b>0$)



      Particularly useful here is:



      $$sqrt{a^2b}=asqrt b$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Use that:



        $$sqrt{ab}=sqrt asqrt b$$
        (when $a,b>0$)



        Particularly useful here is:



        $$sqrt{a^2b}=asqrt b$$






        share|cite|improve this answer









        $endgroup$



        Use that:



        $$sqrt{ab}=sqrt asqrt b$$
        (when $a,b>0$)



        Particularly useful here is:



        $$sqrt{a^2b}=asqrt b$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 16:41









        Rhys HughesRhys Hughes

        7,0501630




        7,0501630






























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