Rationalizing denominator of $frac{18}{sqrt{162}}$. Cannot match textbook solution












2












$begingroup$


I am given this expression and asked to simplify by rationalizing the denominator:



$$frac{18}{sqrt{162}}$$



The solution is provided:



$sqrt{2}$



I arrived at:



$$frac{sqrt{162}}{9}$$



Here is my thought process to arrive at this incorrect answer:



$frac{18}{sqrt{162}}$



= $frac{18}{sqrt{162}}$ * $frac{sqrt{162}}{sqrt{162}}$



= $frac{18sqrt{162}}{162}$



= $frac{sqrt{162}}{9}$



How can I arrive at $sqrt{2}$ ?






algebra-precalculus







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    Hint: $162=2cdot 81$.
    $endgroup$
    – Michael Burr
    Jan 3 at 4:52








  • 1




    $begingroup$
    $162=2*81=2*9^2$ so $sqrt {162}=sqrt {2*9^2}=9sqrt 2$. If your hadn't "deradicalized" the denominator you would have ended up with $frac 2 {sqrt 2} $ which is also deradicalized as $sqrt 2$.
    $endgroup$
    – fleablood
    Jan 3 at 5:49
















2












$begingroup$


I am given this expression and asked to simplify by rationalizing the denominator:



$$frac{18}{sqrt{162}}$$



The solution is provided:



$sqrt{2}$



I arrived at:



$$frac{sqrt{162}}{9}$$



Here is my thought process to arrive at this incorrect answer:



$frac{18}{sqrt{162}}$



= $frac{18}{sqrt{162}}$ * $frac{sqrt{162}}{sqrt{162}}$



= $frac{18sqrt{162}}{162}$



= $frac{sqrt{162}}{9}$



How can I arrive at $sqrt{2}$ ?






algebra-precalculus







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    Hint: $162=2cdot 81$.
    $endgroup$
    – Michael Burr
    Jan 3 at 4:52








  • 1




    $begingroup$
    $162=2*81=2*9^2$ so $sqrt {162}=sqrt {2*9^2}=9sqrt 2$. If your hadn't "deradicalized" the denominator you would have ended up with $frac 2 {sqrt 2} $ which is also deradicalized as $sqrt 2$.
    $endgroup$
    – fleablood
    Jan 3 at 5:49














2












2








2





$begingroup$


I am given this expression and asked to simplify by rationalizing the denominator:



$$frac{18}{sqrt{162}}$$



The solution is provided:



$sqrt{2}$



I arrived at:



$$frac{sqrt{162}}{9}$$



Here is my thought process to arrive at this incorrect answer:



$frac{18}{sqrt{162}}$



= $frac{18}{sqrt{162}}$ * $frac{sqrt{162}}{sqrt{162}}$



= $frac{18sqrt{162}}{162}$



= $frac{sqrt{162}}{9}$



How can I arrive at $sqrt{2}$ ?






algebra-precalculus







share|cite|improve this question









$endgroup$




I am given this expression and asked to simplify by rationalizing the denominator:



$$frac{18}{sqrt{162}}$$



The solution is provided:



$sqrt{2}$



I arrived at:



$$frac{sqrt{162}}{9}$$



Here is my thought process to arrive at this incorrect answer:



$frac{18}{sqrt{162}}$



= $frac{18}{sqrt{162}}$ * $frac{sqrt{162}}{sqrt{162}}$



= $frac{18sqrt{162}}{162}$



= $frac{sqrt{162}}{9}$



How can I arrive at $sqrt{2}$ ?






algebra-precalculus




algebra-precalculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 3 at 4:47









Doug FirDoug Fir

3227




3227








  • 4




    $begingroup$
    Hint: $162=2cdot 81$.
    $endgroup$
    – Michael Burr
    Jan 3 at 4:52








  • 1




    $begingroup$
    $162=2*81=2*9^2$ so $sqrt {162}=sqrt {2*9^2}=9sqrt 2$. If your hadn't "deradicalized" the denominator you would have ended up with $frac 2 {sqrt 2} $ which is also deradicalized as $sqrt 2$.
    $endgroup$
    – fleablood
    Jan 3 at 5:49














  • 4




    $begingroup$
    Hint: $162=2cdot 81$.
    $endgroup$
    – Michael Burr
    Jan 3 at 4:52








  • 1




    $begingroup$
    $162=2*81=2*9^2$ so $sqrt {162}=sqrt {2*9^2}=9sqrt 2$. If your hadn't "deradicalized" the denominator you would have ended up with $frac 2 {sqrt 2} $ which is also deradicalized as $sqrt 2$.
    $endgroup$
    – fleablood
    Jan 3 at 5:49








4




4




$begingroup$
Hint: $162=2cdot 81$.
$endgroup$
– Michael Burr
Jan 3 at 4:52






$begingroup$
Hint: $162=2cdot 81$.
$endgroup$
– Michael Burr
Jan 3 at 4:52






1




1




$begingroup$
$162=2*81=2*9^2$ so $sqrt {162}=sqrt {2*9^2}=9sqrt 2$. If your hadn't "deradicalized" the denominator you would have ended up with $frac 2 {sqrt 2} $ which is also deradicalized as $sqrt 2$.
$endgroup$
– fleablood
Jan 3 at 5:49




$begingroup$
$162=2*81=2*9^2$ so $sqrt {162}=sqrt {2*9^2}=9sqrt 2$. If your hadn't "deradicalized" the denominator you would have ended up with $frac 2 {sqrt 2} $ which is also deradicalized as $sqrt 2$.
$endgroup$
– fleablood
Jan 3 at 5:49










4 Answers
4






active

oldest

votes


















7












$begingroup$

$frac{sqrt{162}}{9} = frac{sqrt{2 cdot 9^2}}{9} = frac{9sqrt{2}}{9} = sqrt{2}$.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    $$require{cancel}frac{18}{sqrt{162}}=frac{2cdot3^2}{sqrt{2cdot3^4}}=frac{2cdotcancel{3^2}}{sqrt{2}cdotcancel{3^2}}=frac{2}{sqrt{2}}cdotfrac{sqrt{2}}{sqrt{2}}=frac{cancel2sqrt{2}}{cancel{2}}=sqrt{2}$$






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      Your thought process is good. But just continue with factorizing $162=2*81=2*3^4$.



      So $sqrt {162}=sqrt {2*3^4}=sqrt {2}sqrt {3^4}=sqrt 2*3^2=9sqrt 2$ and from there.... it's just mechanics.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        $ sqrt{162} $ needs to be simplified further, as $ 162 $ is the product containing a perfect square (i.e. $ 81 $). Thus $$ sqrt{162} = sqrt{2 cdot 81} = sqrt{2}sqrt{81} = 9sqrt{2}$$



        and hence $$ frac{sqrt{162}}{9} = frac{9sqrt{2}}{9} = sqrt{2} $$






        share|cite|improve this answer









        $endgroup$













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          4 Answers
          4






          active

          oldest

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          4 Answers
          4






          active

          oldest

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          active

          oldest

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          active

          oldest

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          7












          $begingroup$

          $frac{sqrt{162}}{9} = frac{sqrt{2 cdot 9^2}}{9} = frac{9sqrt{2}}{9} = sqrt{2}$.






          share|cite|improve this answer









          $endgroup$


















            7












            $begingroup$

            $frac{sqrt{162}}{9} = frac{sqrt{2 cdot 9^2}}{9} = frac{9sqrt{2}}{9} = sqrt{2}$.






            share|cite|improve this answer









            $endgroup$
















              7












              7








              7





              $begingroup$

              $frac{sqrt{162}}{9} = frac{sqrt{2 cdot 9^2}}{9} = frac{9sqrt{2}}{9} = sqrt{2}$.






              share|cite|improve this answer









              $endgroup$



              $frac{sqrt{162}}{9} = frac{sqrt{2 cdot 9^2}}{9} = frac{9sqrt{2}}{9} = sqrt{2}$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 3 at 4:53









              mlerma54mlerma54

              1,177148




              1,177148























                  3












                  $begingroup$

                  $$require{cancel}frac{18}{sqrt{162}}=frac{2cdot3^2}{sqrt{2cdot3^4}}=frac{2cdotcancel{3^2}}{sqrt{2}cdotcancel{3^2}}=frac{2}{sqrt{2}}cdotfrac{sqrt{2}}{sqrt{2}}=frac{cancel2sqrt{2}}{cancel{2}}=sqrt{2}$$






                  share|cite|improve this answer









                  $endgroup$


















                    3












                    $begingroup$

                    $$require{cancel}frac{18}{sqrt{162}}=frac{2cdot3^2}{sqrt{2cdot3^4}}=frac{2cdotcancel{3^2}}{sqrt{2}cdotcancel{3^2}}=frac{2}{sqrt{2}}cdotfrac{sqrt{2}}{sqrt{2}}=frac{cancel2sqrt{2}}{cancel{2}}=sqrt{2}$$






                    share|cite|improve this answer









                    $endgroup$
















                      3












                      3








                      3





                      $begingroup$

                      $$require{cancel}frac{18}{sqrt{162}}=frac{2cdot3^2}{sqrt{2cdot3^4}}=frac{2cdotcancel{3^2}}{sqrt{2}cdotcancel{3^2}}=frac{2}{sqrt{2}}cdotfrac{sqrt{2}}{sqrt{2}}=frac{cancel2sqrt{2}}{cancel{2}}=sqrt{2}$$






                      share|cite|improve this answer









                      $endgroup$



                      $$require{cancel}frac{18}{sqrt{162}}=frac{2cdot3^2}{sqrt{2cdot3^4}}=frac{2cdotcancel{3^2}}{sqrt{2}cdotcancel{3^2}}=frac{2}{sqrt{2}}cdotfrac{sqrt{2}}{sqrt{2}}=frac{cancel2sqrt{2}}{cancel{2}}=sqrt{2}$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 3 at 4:56









                      John GlennJohn Glenn

                      1,957424




                      1,957424























                          2












                          $begingroup$

                          Your thought process is good. But just continue with factorizing $162=2*81=2*3^4$.



                          So $sqrt {162}=sqrt {2*3^4}=sqrt {2}sqrt {3^4}=sqrt 2*3^2=9sqrt 2$ and from there.... it's just mechanics.






                          share|cite|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            Your thought process is good. But just continue with factorizing $162=2*81=2*3^4$.



                            So $sqrt {162}=sqrt {2*3^4}=sqrt {2}sqrt {3^4}=sqrt 2*3^2=9sqrt 2$ and from there.... it's just mechanics.






                            share|cite|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              Your thought process is good. But just continue with factorizing $162=2*81=2*3^4$.



                              So $sqrt {162}=sqrt {2*3^4}=sqrt {2}sqrt {3^4}=sqrt 2*3^2=9sqrt 2$ and from there.... it's just mechanics.






                              share|cite|improve this answer









                              $endgroup$



                              Your thought process is good. But just continue with factorizing $162=2*81=2*3^4$.



                              So $sqrt {162}=sqrt {2*3^4}=sqrt {2}sqrt {3^4}=sqrt 2*3^2=9sqrt 2$ and from there.... it's just mechanics.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 3 at 5:54









                              fleabloodfleablood

                              69.5k22685




                              69.5k22685























                                  0












                                  $begingroup$

                                  $ sqrt{162} $ needs to be simplified further, as $ 162 $ is the product containing a perfect square (i.e. $ 81 $). Thus $$ sqrt{162} = sqrt{2 cdot 81} = sqrt{2}sqrt{81} = 9sqrt{2}$$



                                  and hence $$ frac{sqrt{162}}{9} = frac{9sqrt{2}}{9} = sqrt{2} $$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    $ sqrt{162} $ needs to be simplified further, as $ 162 $ is the product containing a perfect square (i.e. $ 81 $). Thus $$ sqrt{162} = sqrt{2 cdot 81} = sqrt{2}sqrt{81} = 9sqrt{2}$$



                                    and hence $$ frac{sqrt{162}}{9} = frac{9sqrt{2}}{9} = sqrt{2} $$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      $ sqrt{162} $ needs to be simplified further, as $ 162 $ is the product containing a perfect square (i.e. $ 81 $). Thus $$ sqrt{162} = sqrt{2 cdot 81} = sqrt{2}sqrt{81} = 9sqrt{2}$$



                                      and hence $$ frac{sqrt{162}}{9} = frac{9sqrt{2}}{9} = sqrt{2} $$






                                      share|cite|improve this answer









                                      $endgroup$



                                      $ sqrt{162} $ needs to be simplified further, as $ 162 $ is the product containing a perfect square (i.e. $ 81 $). Thus $$ sqrt{162} = sqrt{2 cdot 81} = sqrt{2}sqrt{81} = 9sqrt{2}$$



                                      and hence $$ frac{sqrt{162}}{9} = frac{9sqrt{2}}{9} = sqrt{2} $$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 17 at 18:05









                                      Marvin CohenMarvin Cohen

                                      131117




                                      131117






























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