Trying to simplify $frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2}$ into $frac{-5sqrt{2}-6}{7}$












4












$begingroup$


I'm asked to simplify $frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2}$ and am provided with the solution $frac{-5sqrt{2}-6}{7}$



I have tried several approaches and failed. Here's one path I took:



(Will try to simplify the left hand side fraction part first and then deal with the $-2^{1/2}$ later)



$frac{sqrt{8}-sqrt{16}}{4-sqrt{2}}$
The root of 16 is 4 and the root of 8 could be written as $2sqrt{2}$ thus:



$frac{2sqrt{2}-4}{4-sqrt{2}}$



Not really sure where to go from here so I tried multiplying out the radical in the denominator:



$frac{2sqrt{2}-4}{4-sqrt{2}}$ = $frac{2sqrt{2}-4}{4-sqrt{2}} * frac{4+sqrt{2}}{4+sqrt{2}}$ = $frac{(2sqrt{2}-4)(4+sqrt{2})}{16-2}$ =



(I become less certain in my working here)



$frac{8sqrt{2}*2(sqrt{2}^2)-16-4sqrt{2}}{14}$ = $frac{8sqrt{2}*4-16-4sqrt{2}}{14}$ = $frac{32sqrt{2}-16-4sqrt{2}}{14}$ = $frac{28sqrt{2}-16}{14}$



Then add back the $-2^{1/2}$ which can also be written as $sqrt{2}$



This is as far as I can get. I don't know if $frac{28sqrt{2}-16}{14}-sqrt{2}$ is still correct or close to the solution. How can I arrive at $frac{-5sqrt{2}-6}{7}$?






algebra-precalculus radicals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Did you mean $8sqrt{2} ast 2 ast (sqrt{2})^2$? I think you should have two terms here
    $endgroup$
    – bounceback
    Jan 7 at 2:48
















4












$begingroup$


I'm asked to simplify $frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2}$ and am provided with the solution $frac{-5sqrt{2}-6}{7}$



I have tried several approaches and failed. Here's one path I took:



(Will try to simplify the left hand side fraction part first and then deal with the $-2^{1/2}$ later)



$frac{sqrt{8}-sqrt{16}}{4-sqrt{2}}$
The root of 16 is 4 and the root of 8 could be written as $2sqrt{2}$ thus:



$frac{2sqrt{2}-4}{4-sqrt{2}}$



Not really sure where to go from here so I tried multiplying out the radical in the denominator:



$frac{2sqrt{2}-4}{4-sqrt{2}}$ = $frac{2sqrt{2}-4}{4-sqrt{2}} * frac{4+sqrt{2}}{4+sqrt{2}}$ = $frac{(2sqrt{2}-4)(4+sqrt{2})}{16-2}$ =



(I become less certain in my working here)



$frac{8sqrt{2}*2(sqrt{2}^2)-16-4sqrt{2}}{14}$ = $frac{8sqrt{2}*4-16-4sqrt{2}}{14}$ = $frac{32sqrt{2}-16-4sqrt{2}}{14}$ = $frac{28sqrt{2}-16}{14}$



Then add back the $-2^{1/2}$ which can also be written as $sqrt{2}$



This is as far as I can get. I don't know if $frac{28sqrt{2}-16}{14}-sqrt{2}$ is still correct or close to the solution. How can I arrive at $frac{-5sqrt{2}-6}{7}$?






algebra-precalculus radicals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Did you mean $8sqrt{2} ast 2 ast (sqrt{2})^2$? I think you should have two terms here
    $endgroup$
    – bounceback
    Jan 7 at 2:48














4












4








4





$begingroup$


I'm asked to simplify $frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2}$ and am provided with the solution $frac{-5sqrt{2}-6}{7}$



I have tried several approaches and failed. Here's one path I took:



(Will try to simplify the left hand side fraction part first and then deal with the $-2^{1/2}$ later)



$frac{sqrt{8}-sqrt{16}}{4-sqrt{2}}$
The root of 16 is 4 and the root of 8 could be written as $2sqrt{2}$ thus:



$frac{2sqrt{2}-4}{4-sqrt{2}}$



Not really sure where to go from here so I tried multiplying out the radical in the denominator:



$frac{2sqrt{2}-4}{4-sqrt{2}}$ = $frac{2sqrt{2}-4}{4-sqrt{2}} * frac{4+sqrt{2}}{4+sqrt{2}}$ = $frac{(2sqrt{2}-4)(4+sqrt{2})}{16-2}$ =



(I become less certain in my working here)



$frac{8sqrt{2}*2(sqrt{2}^2)-16-4sqrt{2}}{14}$ = $frac{8sqrt{2}*4-16-4sqrt{2}}{14}$ = $frac{32sqrt{2}-16-4sqrt{2}}{14}$ = $frac{28sqrt{2}-16}{14}$



Then add back the $-2^{1/2}$ which can also be written as $sqrt{2}$



This is as far as I can get. I don't know if $frac{28sqrt{2}-16}{14}-sqrt{2}$ is still correct or close to the solution. How can I arrive at $frac{-5sqrt{2}-6}{7}$?






algebra-precalculus radicals







share|cite|improve this question











$endgroup$




I'm asked to simplify $frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2}$ and am provided with the solution $frac{-5sqrt{2}-6}{7}$



I have tried several approaches and failed. Here's one path I took:



(Will try to simplify the left hand side fraction part first and then deal with the $-2^{1/2}$ later)



$frac{sqrt{8}-sqrt{16}}{4-sqrt{2}}$
The root of 16 is 4 and the root of 8 could be written as $2sqrt{2}$ thus:



$frac{2sqrt{2}-4}{4-sqrt{2}}$



Not really sure where to go from here so I tried multiplying out the radical in the denominator:



$frac{2sqrt{2}-4}{4-sqrt{2}}$ = $frac{2sqrt{2}-4}{4-sqrt{2}} * frac{4+sqrt{2}}{4+sqrt{2}}$ = $frac{(2sqrt{2}-4)(4+sqrt{2})}{16-2}$ =



(I become less certain in my working here)



$frac{8sqrt{2}*2(sqrt{2}^2)-16-4sqrt{2}}{14}$ = $frac{8sqrt{2}*4-16-4sqrt{2}}{14}$ = $frac{32sqrt{2}-16-4sqrt{2}}{14}$ = $frac{28sqrt{2}-16}{14}$



Then add back the $-2^{1/2}$ which can also be written as $sqrt{2}$



This is as far as I can get. I don't know if $frac{28sqrt{2}-16}{14}-sqrt{2}$ is still correct or close to the solution. How can I arrive at $frac{-5sqrt{2}-6}{7}$?






algebra-precalculus radicals




algebra-precalculus radicals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 10:22









Martin Sleziak

44.7k9117272




44.7k9117272










asked Jan 7 at 2:44









Doug FirDoug Fir

3227




3227








  • 1




    $begingroup$
    Did you mean $8sqrt{2} ast 2 ast (sqrt{2})^2$? I think you should have two terms here
    $endgroup$
    – bounceback
    Jan 7 at 2:48














  • 1




    $begingroup$
    Did you mean $8sqrt{2} ast 2 ast (sqrt{2})^2$? I think you should have two terms here
    $endgroup$
    – bounceback
    Jan 7 at 2:48








1




1




$begingroup$
Did you mean $8sqrt{2} ast 2 ast (sqrt{2})^2$? I think you should have two terms here
$endgroup$
– bounceback
Jan 7 at 2:48




$begingroup$
Did you mean $8sqrt{2} ast 2 ast (sqrt{2})^2$? I think you should have two terms here
$endgroup$
– bounceback
Jan 7 at 2:48










3 Answers
3






active

oldest

votes


















3












$begingroup$

You were doing fine until the place where you tried to expand
$(2sqrt2 - 4)(4 + sqrt2).$



There are mnemonic techniques for this but I think plain old distributive law works well enough:
begin{align}
(2sqrt2 - 4)(4 + sqrt2) &= (2sqrt2 - 4)4 + (2sqrt2 - 4)sqrt2 \
&= (8sqrt2 - 16) + (4 - 4sqrt2) \
&= 4sqrt2 - 12.
end{align}



Next you might notice a chance to cancel a factor of $2$ in the numerator and denominator of
$frac{4sqrt2 - 12}{14}.$



And finally you'll want to change the $-sqrt2$ so that you have two fractions with a common denominator and can finish.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hi David. Regarding your last sentence, would it be possible to spell that out for me? What's the rule here?
    $endgroup$
    – Doug Fir
    Jan 7 at 3:38






  • 1




    $begingroup$
    It's the same rule you would apply to simplify something like $frac37 - 2.$ The $2$ is equal to $frac21,$ which is equal to $frac{7cdot2}{7cdot1}.$ In your problem you have $sqrt2$ instead of $2$ but the principle is the same.
    $endgroup$
    – David K
    Jan 7 at 3:42










  • $begingroup$
    Hi David, thanks for clarifying that, I understand it now
    $endgroup$
    – Doug Fir
    Jan 7 at 4:01



















4












$begingroup$

$$begin{align}
frac{sqrt{8}-sqrt{16}}{4-sqrt{2}}-sqrt{2}&=frac{2sqrt{2}-4}{4-sqrt{2}}-sqrt{2}\
&=frac{2sqrt{2}-4}{4-sqrt{2}}-frac{4sqrt{2}-2}{4-sqrt{2}}\
&=frac{-2sqrt{2}-2}{4-sqrt{2}}\
&=frac{-2sqrt{2}-2}{4-sqrt{2}}~cdot~frac{4+sqrt{2}}{4+sqrt{2}}\
&=frac{-10sqrt{2}-12}{14}\
&=frac{-5sqrt{2}-6}{7}
end{align}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for the answer. I'm trying to understand what you are doing on the first new line, where you subtract the fraction $frac{4sqrt{2}-2}{4sqrt{2}-2}$. 1. What's the objective here and 2. Why is the new denominator unchanged, since the next lines denominator is the same, $4-sqrt{2}$?
    $endgroup$
    – Doug Fir
    Jan 7 at 3:12



















3












$begingroup$

begin{align}
frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2} & = frac{2sqrt{2}-4}{4-sqrt{2}}cdot frac{4+sqrt{2}}{4+sqrt{2}} - sqrt{2} \
& = frac{4sqrt{2}-12}{14} - sqrt{2} \
& = frac{2sqrt{2}-6}{7} - sqrt{2} \
& = frac{2sqrt{2}-6 -7 sqrt{2}}{7}\
& = frac{-5sqrt{2} -6 }{7}
end{align}






share|cite|improve this answer









$endgroup$













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    You were doing fine until the place where you tried to expand
    $(2sqrt2 - 4)(4 + sqrt2).$



    There are mnemonic techniques for this but I think plain old distributive law works well enough:
    begin{align}
    (2sqrt2 - 4)(4 + sqrt2) &= (2sqrt2 - 4)4 + (2sqrt2 - 4)sqrt2 \
    &= (8sqrt2 - 16) + (4 - 4sqrt2) \
    &= 4sqrt2 - 12.
    end{align}



    Next you might notice a chance to cancel a factor of $2$ in the numerator and denominator of
    $frac{4sqrt2 - 12}{14}.$



    And finally you'll want to change the $-sqrt2$ so that you have two fractions with a common denominator and can finish.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Hi David. Regarding your last sentence, would it be possible to spell that out for me? What's the rule here?
      $endgroup$
      – Doug Fir
      Jan 7 at 3:38






    • 1




      $begingroup$
      It's the same rule you would apply to simplify something like $frac37 - 2.$ The $2$ is equal to $frac21,$ which is equal to $frac{7cdot2}{7cdot1}.$ In your problem you have $sqrt2$ instead of $2$ but the principle is the same.
      $endgroup$
      – David K
      Jan 7 at 3:42










    • $begingroup$
      Hi David, thanks for clarifying that, I understand it now
      $endgroup$
      – Doug Fir
      Jan 7 at 4:01
















    3












    $begingroup$

    You were doing fine until the place where you tried to expand
    $(2sqrt2 - 4)(4 + sqrt2).$



    There are mnemonic techniques for this but I think plain old distributive law works well enough:
    begin{align}
    (2sqrt2 - 4)(4 + sqrt2) &= (2sqrt2 - 4)4 + (2sqrt2 - 4)sqrt2 \
    &= (8sqrt2 - 16) + (4 - 4sqrt2) \
    &= 4sqrt2 - 12.
    end{align}



    Next you might notice a chance to cancel a factor of $2$ in the numerator and denominator of
    $frac{4sqrt2 - 12}{14}.$



    And finally you'll want to change the $-sqrt2$ so that you have two fractions with a common denominator and can finish.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Hi David. Regarding your last sentence, would it be possible to spell that out for me? What's the rule here?
      $endgroup$
      – Doug Fir
      Jan 7 at 3:38






    • 1




      $begingroup$
      It's the same rule you would apply to simplify something like $frac37 - 2.$ The $2$ is equal to $frac21,$ which is equal to $frac{7cdot2}{7cdot1}.$ In your problem you have $sqrt2$ instead of $2$ but the principle is the same.
      $endgroup$
      – David K
      Jan 7 at 3:42










    • $begingroup$
      Hi David, thanks for clarifying that, I understand it now
      $endgroup$
      – Doug Fir
      Jan 7 at 4:01














    3












    3








    3





    $begingroup$

    You were doing fine until the place where you tried to expand
    $(2sqrt2 - 4)(4 + sqrt2).$



    There are mnemonic techniques for this but I think plain old distributive law works well enough:
    begin{align}
    (2sqrt2 - 4)(4 + sqrt2) &= (2sqrt2 - 4)4 + (2sqrt2 - 4)sqrt2 \
    &= (8sqrt2 - 16) + (4 - 4sqrt2) \
    &= 4sqrt2 - 12.
    end{align}



    Next you might notice a chance to cancel a factor of $2$ in the numerator and denominator of
    $frac{4sqrt2 - 12}{14}.$



    And finally you'll want to change the $-sqrt2$ so that you have two fractions with a common denominator and can finish.






    share|cite|improve this answer









    $endgroup$



    You were doing fine until the place where you tried to expand
    $(2sqrt2 - 4)(4 + sqrt2).$



    There are mnemonic techniques for this but I think plain old distributive law works well enough:
    begin{align}
    (2sqrt2 - 4)(4 + sqrt2) &= (2sqrt2 - 4)4 + (2sqrt2 - 4)sqrt2 \
    &= (8sqrt2 - 16) + (4 - 4sqrt2) \
    &= 4sqrt2 - 12.
    end{align}



    Next you might notice a chance to cancel a factor of $2$ in the numerator and denominator of
    $frac{4sqrt2 - 12}{14}.$



    And finally you'll want to change the $-sqrt2$ so that you have two fractions with a common denominator and can finish.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 7 at 3:20









    David KDavid K

    53.8k342116




    53.8k342116












    • $begingroup$
      Hi David. Regarding your last sentence, would it be possible to spell that out for me? What's the rule here?
      $endgroup$
      – Doug Fir
      Jan 7 at 3:38






    • 1




      $begingroup$
      It's the same rule you would apply to simplify something like $frac37 - 2.$ The $2$ is equal to $frac21,$ which is equal to $frac{7cdot2}{7cdot1}.$ In your problem you have $sqrt2$ instead of $2$ but the principle is the same.
      $endgroup$
      – David K
      Jan 7 at 3:42










    • $begingroup$
      Hi David, thanks for clarifying that, I understand it now
      $endgroup$
      – Doug Fir
      Jan 7 at 4:01


















    • $begingroup$
      Hi David. Regarding your last sentence, would it be possible to spell that out for me? What's the rule here?
      $endgroup$
      – Doug Fir
      Jan 7 at 3:38






    • 1




      $begingroup$
      It's the same rule you would apply to simplify something like $frac37 - 2.$ The $2$ is equal to $frac21,$ which is equal to $frac{7cdot2}{7cdot1}.$ In your problem you have $sqrt2$ instead of $2$ but the principle is the same.
      $endgroup$
      – David K
      Jan 7 at 3:42










    • $begingroup$
      Hi David, thanks for clarifying that, I understand it now
      $endgroup$
      – Doug Fir
      Jan 7 at 4:01
















    $begingroup$
    Hi David. Regarding your last sentence, would it be possible to spell that out for me? What's the rule here?
    $endgroup$
    – Doug Fir
    Jan 7 at 3:38




    $begingroup$
    Hi David. Regarding your last sentence, would it be possible to spell that out for me? What's the rule here?
    $endgroup$
    – Doug Fir
    Jan 7 at 3:38




    1




    1




    $begingroup$
    It's the same rule you would apply to simplify something like $frac37 - 2.$ The $2$ is equal to $frac21,$ which is equal to $frac{7cdot2}{7cdot1}.$ In your problem you have $sqrt2$ instead of $2$ but the principle is the same.
    $endgroup$
    – David K
    Jan 7 at 3:42




    $begingroup$
    It's the same rule you would apply to simplify something like $frac37 - 2.$ The $2$ is equal to $frac21,$ which is equal to $frac{7cdot2}{7cdot1}.$ In your problem you have $sqrt2$ instead of $2$ but the principle is the same.
    $endgroup$
    – David K
    Jan 7 at 3:42












    $begingroup$
    Hi David, thanks for clarifying that, I understand it now
    $endgroup$
    – Doug Fir
    Jan 7 at 4:01




    $begingroup$
    Hi David, thanks for clarifying that, I understand it now
    $endgroup$
    – Doug Fir
    Jan 7 at 4:01











    4












    $begingroup$

    $$begin{align}
    frac{sqrt{8}-sqrt{16}}{4-sqrt{2}}-sqrt{2}&=frac{2sqrt{2}-4}{4-sqrt{2}}-sqrt{2}\
    &=frac{2sqrt{2}-4}{4-sqrt{2}}-frac{4sqrt{2}-2}{4-sqrt{2}}\
    &=frac{-2sqrt{2}-2}{4-sqrt{2}}\
    &=frac{-2sqrt{2}-2}{4-sqrt{2}}~cdot~frac{4+sqrt{2}}{4+sqrt{2}}\
    &=frac{-10sqrt{2}-12}{14}\
    &=frac{-5sqrt{2}-6}{7}
    end{align}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for the answer. I'm trying to understand what you are doing on the first new line, where you subtract the fraction $frac{4sqrt{2}-2}{4sqrt{2}-2}$. 1. What's the objective here and 2. Why is the new denominator unchanged, since the next lines denominator is the same, $4-sqrt{2}$?
      $endgroup$
      – Doug Fir
      Jan 7 at 3:12
















    4












    $begingroup$

    $$begin{align}
    frac{sqrt{8}-sqrt{16}}{4-sqrt{2}}-sqrt{2}&=frac{2sqrt{2}-4}{4-sqrt{2}}-sqrt{2}\
    &=frac{2sqrt{2}-4}{4-sqrt{2}}-frac{4sqrt{2}-2}{4-sqrt{2}}\
    &=frac{-2sqrt{2}-2}{4-sqrt{2}}\
    &=frac{-2sqrt{2}-2}{4-sqrt{2}}~cdot~frac{4+sqrt{2}}{4+sqrt{2}}\
    &=frac{-10sqrt{2}-12}{14}\
    &=frac{-5sqrt{2}-6}{7}
    end{align}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for the answer. I'm trying to understand what you are doing on the first new line, where you subtract the fraction $frac{4sqrt{2}-2}{4sqrt{2}-2}$. 1. What's the objective here and 2. Why is the new denominator unchanged, since the next lines denominator is the same, $4-sqrt{2}$?
      $endgroup$
      – Doug Fir
      Jan 7 at 3:12














    4












    4








    4





    $begingroup$

    $$begin{align}
    frac{sqrt{8}-sqrt{16}}{4-sqrt{2}}-sqrt{2}&=frac{2sqrt{2}-4}{4-sqrt{2}}-sqrt{2}\
    &=frac{2sqrt{2}-4}{4-sqrt{2}}-frac{4sqrt{2}-2}{4-sqrt{2}}\
    &=frac{-2sqrt{2}-2}{4-sqrt{2}}\
    &=frac{-2sqrt{2}-2}{4-sqrt{2}}~cdot~frac{4+sqrt{2}}{4+sqrt{2}}\
    &=frac{-10sqrt{2}-12}{14}\
    &=frac{-5sqrt{2}-6}{7}
    end{align}$$






    share|cite|improve this answer









    $endgroup$



    $$begin{align}
    frac{sqrt{8}-sqrt{16}}{4-sqrt{2}}-sqrt{2}&=frac{2sqrt{2}-4}{4-sqrt{2}}-sqrt{2}\
    &=frac{2sqrt{2}-4}{4-sqrt{2}}-frac{4sqrt{2}-2}{4-sqrt{2}}\
    &=frac{-2sqrt{2}-2}{4-sqrt{2}}\
    &=frac{-2sqrt{2}-2}{4-sqrt{2}}~cdot~frac{4+sqrt{2}}{4+sqrt{2}}\
    &=frac{-10sqrt{2}-12}{14}\
    &=frac{-5sqrt{2}-6}{7}
    end{align}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 7 at 3:01









    LarryLarry

    2,31431028




    2,31431028












    • $begingroup$
      Thank you for the answer. I'm trying to understand what you are doing on the first new line, where you subtract the fraction $frac{4sqrt{2}-2}{4sqrt{2}-2}$. 1. What's the objective here and 2. Why is the new denominator unchanged, since the next lines denominator is the same, $4-sqrt{2}$?
      $endgroup$
      – Doug Fir
      Jan 7 at 3:12


















    • $begingroup$
      Thank you for the answer. I'm trying to understand what you are doing on the first new line, where you subtract the fraction $frac{4sqrt{2}-2}{4sqrt{2}-2}$. 1. What's the objective here and 2. Why is the new denominator unchanged, since the next lines denominator is the same, $4-sqrt{2}$?
      $endgroup$
      – Doug Fir
      Jan 7 at 3:12
















    $begingroup$
    Thank you for the answer. I'm trying to understand what you are doing on the first new line, where you subtract the fraction $frac{4sqrt{2}-2}{4sqrt{2}-2}$. 1. What's the objective here and 2. Why is the new denominator unchanged, since the next lines denominator is the same, $4-sqrt{2}$?
    $endgroup$
    – Doug Fir
    Jan 7 at 3:12




    $begingroup$
    Thank you for the answer. I'm trying to understand what you are doing on the first new line, where you subtract the fraction $frac{4sqrt{2}-2}{4sqrt{2}-2}$. 1. What's the objective here and 2. Why is the new denominator unchanged, since the next lines denominator is the same, $4-sqrt{2}$?
    $endgroup$
    – Doug Fir
    Jan 7 at 3:12











    3












    $begingroup$

    begin{align}
    frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2} & = frac{2sqrt{2}-4}{4-sqrt{2}}cdot frac{4+sqrt{2}}{4+sqrt{2}} - sqrt{2} \
    & = frac{4sqrt{2}-12}{14} - sqrt{2} \
    & = frac{2sqrt{2}-6}{7} - sqrt{2} \
    & = frac{2sqrt{2}-6 -7 sqrt{2}}{7}\
    & = frac{-5sqrt{2} -6 }{7}
    end{align}






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      begin{align}
      frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2} & = frac{2sqrt{2}-4}{4-sqrt{2}}cdot frac{4+sqrt{2}}{4+sqrt{2}} - sqrt{2} \
      & = frac{4sqrt{2}-12}{14} - sqrt{2} \
      & = frac{2sqrt{2}-6}{7} - sqrt{2} \
      & = frac{2sqrt{2}-6 -7 sqrt{2}}{7}\
      & = frac{-5sqrt{2} -6 }{7}
      end{align}






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        begin{align}
        frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2} & = frac{2sqrt{2}-4}{4-sqrt{2}}cdot frac{4+sqrt{2}}{4+sqrt{2}} - sqrt{2} \
        & = frac{4sqrt{2}-12}{14} - sqrt{2} \
        & = frac{2sqrt{2}-6}{7} - sqrt{2} \
        & = frac{2sqrt{2}-6 -7 sqrt{2}}{7}\
        & = frac{-5sqrt{2} -6 }{7}
        end{align}






        share|cite|improve this answer









        $endgroup$



        begin{align}
        frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2} & = frac{2sqrt{2}-4}{4-sqrt{2}}cdot frac{4+sqrt{2}}{4+sqrt{2}} - sqrt{2} \
        & = frac{4sqrt{2}-12}{14} - sqrt{2} \
        & = frac{2sqrt{2}-6}{7} - sqrt{2} \
        & = frac{2sqrt{2}-6 -7 sqrt{2}}{7}\
        & = frac{-5sqrt{2} -6 }{7}
        end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 9:08









        kelalakakelalaka

        3261212




        3261212






























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