Struggling to simplify $w^{3/2}sqrt{32} - w^{3/2}sqrt{50}$ to $-wsqrt{2w}$












1












$begingroup$


I'm asked to simplify $w^{3/2}sqrt{32} - w^{3/2}sqrt{50}$ and am provided with the solution: $-wsqrt{2w}$



I arrived at $9sqrt{2}$ but I think I'm confused in understanding communitive rule here.



Here is my working:



$w^{3/2}sqrt{32} - w^{3/2}sqrt{50}$ = $sqrt{w^3}sqrt{32}$ - $sqrt{w^3}sqrt{50}$ # is this correct approach? I made the radical exponent a radical



Then:



$sqrt{32}$ = $sqrt{4}$ * $sqrt{4}$ * $sqrt{2}$ = $2 * 2 * sqrt{2}$ = $4sqrt{2}$



$sqrt{50}$ = $sqrt{2}$ * $sqrt{25}$ = $5sqrt{2}$



So:



$sqrt{w^3}$$4sqrt{2}$ - $sqrt{w^3}5sqrt{2}$ # should the expressions on either side of the minus sign be considered a single factor? i.e. could I also write as ($sqrt{w^3}$$4sqrt{2}$) - ($sqrt{w^3}5sqrt{2}$) )?



Then I'm less sure about where to go next. Since I have a positive $sqrt{w^3}$ and a negative $sqrt{w^3}$ I cancelled those out and was thus left with $9sqrt{2}$.



More generally I was not sure of how to approach this and could not fin a justification for taking the path that I did.



How can I arrive at $-wsqrt{2w}$ per the text book's solution?






algebra-precalculus







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I'm asked to simplify $w^{3/2}sqrt{32} - w^{3/2}sqrt{50}$ and am provided with the solution: $-wsqrt{2w}$



    I arrived at $9sqrt{2}$ but I think I'm confused in understanding communitive rule here.



    Here is my working:



    $w^{3/2}sqrt{32} - w^{3/2}sqrt{50}$ = $sqrt{w^3}sqrt{32}$ - $sqrt{w^3}sqrt{50}$ # is this correct approach? I made the radical exponent a radical



    Then:



    $sqrt{32}$ = $sqrt{4}$ * $sqrt{4}$ * $sqrt{2}$ = $2 * 2 * sqrt{2}$ = $4sqrt{2}$



    $sqrt{50}$ = $sqrt{2}$ * $sqrt{25}$ = $5sqrt{2}$



    So:



    $sqrt{w^3}$$4sqrt{2}$ - $sqrt{w^3}5sqrt{2}$ # should the expressions on either side of the minus sign be considered a single factor? i.e. could I also write as ($sqrt{w^3}$$4sqrt{2}$) - ($sqrt{w^3}5sqrt{2}$) )?



    Then I'm less sure about where to go next. Since I have a positive $sqrt{w^3}$ and a negative $sqrt{w^3}$ I cancelled those out and was thus left with $9sqrt{2}$.



    More generally I was not sure of how to approach this and could not fin a justification for taking the path that I did.



    How can I arrive at $-wsqrt{2w}$ per the text book's solution?






    algebra-precalculus







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I'm asked to simplify $w^{3/2}sqrt{32} - w^{3/2}sqrt{50}$ and am provided with the solution: $-wsqrt{2w}$



      I arrived at $9sqrt{2}$ but I think I'm confused in understanding communitive rule here.



      Here is my working:



      $w^{3/2}sqrt{32} - w^{3/2}sqrt{50}$ = $sqrt{w^3}sqrt{32}$ - $sqrt{w^3}sqrt{50}$ # is this correct approach? I made the radical exponent a radical



      Then:



      $sqrt{32}$ = $sqrt{4}$ * $sqrt{4}$ * $sqrt{2}$ = $2 * 2 * sqrt{2}$ = $4sqrt{2}$



      $sqrt{50}$ = $sqrt{2}$ * $sqrt{25}$ = $5sqrt{2}$



      So:



      $sqrt{w^3}$$4sqrt{2}$ - $sqrt{w^3}5sqrt{2}$ # should the expressions on either side of the minus sign be considered a single factor? i.e. could I also write as ($sqrt{w^3}$$4sqrt{2}$) - ($sqrt{w^3}5sqrt{2}$) )?



      Then I'm less sure about where to go next. Since I have a positive $sqrt{w^3}$ and a negative $sqrt{w^3}$ I cancelled those out and was thus left with $9sqrt{2}$.



      More generally I was not sure of how to approach this and could not fin a justification for taking the path that I did.



      How can I arrive at $-wsqrt{2w}$ per the text book's solution?






      algebra-precalculus







      share|cite|improve this question









      $endgroup$




      I'm asked to simplify $w^{3/2}sqrt{32} - w^{3/2}sqrt{50}$ and am provided with the solution: $-wsqrt{2w}$



      I arrived at $9sqrt{2}$ but I think I'm confused in understanding communitive rule here.



      Here is my working:



      $w^{3/2}sqrt{32} - w^{3/2}sqrt{50}$ = $sqrt{w^3}sqrt{32}$ - $sqrt{w^3}sqrt{50}$ # is this correct approach? I made the radical exponent a radical



      Then:



      $sqrt{32}$ = $sqrt{4}$ * $sqrt{4}$ * $sqrt{2}$ = $2 * 2 * sqrt{2}$ = $4sqrt{2}$



      $sqrt{50}$ = $sqrt{2}$ * $sqrt{25}$ = $5sqrt{2}$



      So:



      $sqrt{w^3}$$4sqrt{2}$ - $sqrt{w^3}5sqrt{2}$ # should the expressions on either side of the minus sign be considered a single factor? i.e. could I also write as ($sqrt{w^3}$$4sqrt{2}$) - ($sqrt{w^3}5sqrt{2}$) )?



      Then I'm less sure about where to go next. Since I have a positive $sqrt{w^3}$ and a negative $sqrt{w^3}$ I cancelled those out and was thus left with $9sqrt{2}$.



      More generally I was not sure of how to approach this and could not fin a justification for taking the path that I did.



      How can I arrive at $-wsqrt{2w}$ per the text book's solution?






      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 6 at 17:44









      Doug FirDoug Fir

      4278




      4278






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Your approach is absolutely right. But note that $$
          begin{align}
          sqrt{w^3}4sqrt{2}- sqrt{w^3}5sqrt{2}&=sqrt{w^3}(4sqrt{2}-5sqrt{2})\
          &=-sqrt{w^3}sqrt{2}=-wsqrt{2w}.
          end{align}$$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thanks for the answer, I follow and understand your solution. How is it that you knew to factor $sqrt{w^3}$? After I had rewritten $sqrt{32}$ as $4sqrt{2}$ and then $sqrt{50}$ as $5sqrt{2}$ I was stumped about next steps. Is this just a practice an intuition thing or is there a prescribed set of rules and order of operations that I'm missing?
            $endgroup$
            – Doug Fir
            Jan 6 at 18:00






          • 2




            $begingroup$
            Both terms have $sqrt{w^3}$ as a factor.
            $endgroup$
            – KM101
            Jan 6 at 18:02






          • 1




            $begingroup$
            Well, I used the distributive property $x (a+b)=xa+xb $. You may find this helpful.
            $endgroup$
            – Thomas Shelby
            Jan 6 at 18:11



















          1












          $begingroup$

          You are on the right track to simplify $ w^frac{3}{2} sqrt{32} - w^frac{3}{2} sqrt{50} $ to as far as



          $$mathrm{(1)} qquad sqrt{w^3} 4 sqrt{2} - sqrt{w^3} 5 sqrt{2} $$



          I would make expression $ (1) $ neater and rewrite as



          $$mathrm{(2)} qquad 4 sqrt{w^3} sqrt{2} - 5 sqrt{w^3} sqrt{2} $$



          These two terms are alike, and combining the two yields



          $$mathrm{(3)} qquad -sqrt{w^3} sqrt{2} $$



          According to one of the properties of radicals, $ sqrt{a} sqrt{b} = sqrt{ab} $. Using that property and rearranging,



          $$mathrm{(4)} qquad -sqrt{2w^3} $$



          Now $ w^3 $ can be expressed as a product involving a perfect square (i.e. $ w^2 $) and a non-perfect square ($ w $). Expression $ (4) $ becomes



          $$mathrm{(5)} qquad -sqrt{2w w^2} $$



          Simplify to get the desired result



          $$ -w sqrt{2w} $$






          share|cite|improve this answer









          $endgroup$














            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Your approach is absolutely right. But note that $$
            begin{align}
            sqrt{w^3}4sqrt{2}- sqrt{w^3}5sqrt{2}&=sqrt{w^3}(4sqrt{2}-5sqrt{2})\
            &=-sqrt{w^3}sqrt{2}=-wsqrt{2w}.
            end{align}$$






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Thanks for the answer, I follow and understand your solution. How is it that you knew to factor $sqrt{w^3}$? After I had rewritten $sqrt{32}$ as $4sqrt{2}$ and then $sqrt{50}$ as $5sqrt{2}$ I was stumped about next steps. Is this just a practice an intuition thing or is there a prescribed set of rules and order of operations that I'm missing?
              $endgroup$
              – Doug Fir
              Jan 6 at 18:00






            • 2




              $begingroup$
              Both terms have $sqrt{w^3}$ as a factor.
              $endgroup$
              – KM101
              Jan 6 at 18:02






            • 1




              $begingroup$
              Well, I used the distributive property $x (a+b)=xa+xb $. You may find this helpful.
              $endgroup$
              – Thomas Shelby
              Jan 6 at 18:11
















            1












            $begingroup$

            Your approach is absolutely right. But note that $$
            begin{align}
            sqrt{w^3}4sqrt{2}- sqrt{w^3}5sqrt{2}&=sqrt{w^3}(4sqrt{2}-5sqrt{2})\
            &=-sqrt{w^3}sqrt{2}=-wsqrt{2w}.
            end{align}$$






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Thanks for the answer, I follow and understand your solution. How is it that you knew to factor $sqrt{w^3}$? After I had rewritten $sqrt{32}$ as $4sqrt{2}$ and then $sqrt{50}$ as $5sqrt{2}$ I was stumped about next steps. Is this just a practice an intuition thing or is there a prescribed set of rules and order of operations that I'm missing?
              $endgroup$
              – Doug Fir
              Jan 6 at 18:00






            • 2




              $begingroup$
              Both terms have $sqrt{w^3}$ as a factor.
              $endgroup$
              – KM101
              Jan 6 at 18:02






            • 1




              $begingroup$
              Well, I used the distributive property $x (a+b)=xa+xb $. You may find this helpful.
              $endgroup$
              – Thomas Shelby
              Jan 6 at 18:11














            1












            1








            1





            $begingroup$

            Your approach is absolutely right. But note that $$
            begin{align}
            sqrt{w^3}4sqrt{2}- sqrt{w^3}5sqrt{2}&=sqrt{w^3}(4sqrt{2}-5sqrt{2})\
            &=-sqrt{w^3}sqrt{2}=-wsqrt{2w}.
            end{align}$$






            share|cite|improve this answer











            $endgroup$



            Your approach is absolutely right. But note that $$
            begin{align}
            sqrt{w^3}4sqrt{2}- sqrt{w^3}5sqrt{2}&=sqrt{w^3}(4sqrt{2}-5sqrt{2})\
            &=-sqrt{w^3}sqrt{2}=-wsqrt{2w}.
            end{align}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 6 at 18:39









            thesmallprint

            2,6951618




            2,6951618










            answered Jan 6 at 17:51









            Thomas ShelbyThomas Shelby

            4,7362727




            4,7362727








            • 1




              $begingroup$
              Thanks for the answer, I follow and understand your solution. How is it that you knew to factor $sqrt{w^3}$? After I had rewritten $sqrt{32}$ as $4sqrt{2}$ and then $sqrt{50}$ as $5sqrt{2}$ I was stumped about next steps. Is this just a practice an intuition thing or is there a prescribed set of rules and order of operations that I'm missing?
              $endgroup$
              – Doug Fir
              Jan 6 at 18:00






            • 2




              $begingroup$
              Both terms have $sqrt{w^3}$ as a factor.
              $endgroup$
              – KM101
              Jan 6 at 18:02






            • 1




              $begingroup$
              Well, I used the distributive property $x (a+b)=xa+xb $. You may find this helpful.
              $endgroup$
              – Thomas Shelby
              Jan 6 at 18:11














            • 1




              $begingroup$
              Thanks for the answer, I follow and understand your solution. How is it that you knew to factor $sqrt{w^3}$? After I had rewritten $sqrt{32}$ as $4sqrt{2}$ and then $sqrt{50}$ as $5sqrt{2}$ I was stumped about next steps. Is this just a practice an intuition thing or is there a prescribed set of rules and order of operations that I'm missing?
              $endgroup$
              – Doug Fir
              Jan 6 at 18:00






            • 2




              $begingroup$
              Both terms have $sqrt{w^3}$ as a factor.
              $endgroup$
              – KM101
              Jan 6 at 18:02






            • 1




              $begingroup$
              Well, I used the distributive property $x (a+b)=xa+xb $. You may find this helpful.
              $endgroup$
              – Thomas Shelby
              Jan 6 at 18:11








            1




            1




            $begingroup$
            Thanks for the answer, I follow and understand your solution. How is it that you knew to factor $sqrt{w^3}$? After I had rewritten $sqrt{32}$ as $4sqrt{2}$ and then $sqrt{50}$ as $5sqrt{2}$ I was stumped about next steps. Is this just a practice an intuition thing or is there a prescribed set of rules and order of operations that I'm missing?
            $endgroup$
            – Doug Fir
            Jan 6 at 18:00




            $begingroup$
            Thanks for the answer, I follow and understand your solution. How is it that you knew to factor $sqrt{w^3}$? After I had rewritten $sqrt{32}$ as $4sqrt{2}$ and then $sqrt{50}$ as $5sqrt{2}$ I was stumped about next steps. Is this just a practice an intuition thing or is there a prescribed set of rules and order of operations that I'm missing?
            $endgroup$
            – Doug Fir
            Jan 6 at 18:00




            2




            2




            $begingroup$
            Both terms have $sqrt{w^3}$ as a factor.
            $endgroup$
            – KM101
            Jan 6 at 18:02




            $begingroup$
            Both terms have $sqrt{w^3}$ as a factor.
            $endgroup$
            – KM101
            Jan 6 at 18:02




            1




            1




            $begingroup$
            Well, I used the distributive property $x (a+b)=xa+xb $. You may find this helpful.
            $endgroup$
            – Thomas Shelby
            Jan 6 at 18:11




            $begingroup$
            Well, I used the distributive property $x (a+b)=xa+xb $. You may find this helpful.
            $endgroup$
            – Thomas Shelby
            Jan 6 at 18:11











            1












            $begingroup$

            You are on the right track to simplify $ w^frac{3}{2} sqrt{32} - w^frac{3}{2} sqrt{50} $ to as far as



            $$mathrm{(1)} qquad sqrt{w^3} 4 sqrt{2} - sqrt{w^3} 5 sqrt{2} $$



            I would make expression $ (1) $ neater and rewrite as



            $$mathrm{(2)} qquad 4 sqrt{w^3} sqrt{2} - 5 sqrt{w^3} sqrt{2} $$



            These two terms are alike, and combining the two yields



            $$mathrm{(3)} qquad -sqrt{w^3} sqrt{2} $$



            According to one of the properties of radicals, $ sqrt{a} sqrt{b} = sqrt{ab} $. Using that property and rearranging,



            $$mathrm{(4)} qquad -sqrt{2w^3} $$



            Now $ w^3 $ can be expressed as a product involving a perfect square (i.e. $ w^2 $) and a non-perfect square ($ w $). Expression $ (4) $ becomes



            $$mathrm{(5)} qquad -sqrt{2w w^2} $$



            Simplify to get the desired result



            $$ -w sqrt{2w} $$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              You are on the right track to simplify $ w^frac{3}{2} sqrt{32} - w^frac{3}{2} sqrt{50} $ to as far as



              $$mathrm{(1)} qquad sqrt{w^3} 4 sqrt{2} - sqrt{w^3} 5 sqrt{2} $$



              I would make expression $ (1) $ neater and rewrite as



              $$mathrm{(2)} qquad 4 sqrt{w^3} sqrt{2} - 5 sqrt{w^3} sqrt{2} $$



              These two terms are alike, and combining the two yields



              $$mathrm{(3)} qquad -sqrt{w^3} sqrt{2} $$



              According to one of the properties of radicals, $ sqrt{a} sqrt{b} = sqrt{ab} $. Using that property and rearranging,



              $$mathrm{(4)} qquad -sqrt{2w^3} $$



              Now $ w^3 $ can be expressed as a product involving a perfect square (i.e. $ w^2 $) and a non-perfect square ($ w $). Expression $ (4) $ becomes



              $$mathrm{(5)} qquad -sqrt{2w w^2} $$



              Simplify to get the desired result



              $$ -w sqrt{2w} $$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                You are on the right track to simplify $ w^frac{3}{2} sqrt{32} - w^frac{3}{2} sqrt{50} $ to as far as



                $$mathrm{(1)} qquad sqrt{w^3} 4 sqrt{2} - sqrt{w^3} 5 sqrt{2} $$



                I would make expression $ (1) $ neater and rewrite as



                $$mathrm{(2)} qquad 4 sqrt{w^3} sqrt{2} - 5 sqrt{w^3} sqrt{2} $$



                These two terms are alike, and combining the two yields



                $$mathrm{(3)} qquad -sqrt{w^3} sqrt{2} $$



                According to one of the properties of radicals, $ sqrt{a} sqrt{b} = sqrt{ab} $. Using that property and rearranging,



                $$mathrm{(4)} qquad -sqrt{2w^3} $$



                Now $ w^3 $ can be expressed as a product involving a perfect square (i.e. $ w^2 $) and a non-perfect square ($ w $). Expression $ (4) $ becomes



                $$mathrm{(5)} qquad -sqrt{2w w^2} $$



                Simplify to get the desired result



                $$ -w sqrt{2w} $$






                share|cite|improve this answer









                $endgroup$



                You are on the right track to simplify $ w^frac{3}{2} sqrt{32} - w^frac{3}{2} sqrt{50} $ to as far as



                $$mathrm{(1)} qquad sqrt{w^3} 4 sqrt{2} - sqrt{w^3} 5 sqrt{2} $$



                I would make expression $ (1) $ neater and rewrite as



                $$mathrm{(2)} qquad 4 sqrt{w^3} sqrt{2} - 5 sqrt{w^3} sqrt{2} $$



                These two terms are alike, and combining the two yields



                $$mathrm{(3)} qquad -sqrt{w^3} sqrt{2} $$



                According to one of the properties of radicals, $ sqrt{a} sqrt{b} = sqrt{ab} $. Using that property and rearranging,



                $$mathrm{(4)} qquad -sqrt{2w^3} $$



                Now $ w^3 $ can be expressed as a product involving a perfect square (i.e. $ w^2 $) and a non-perfect square ($ w $). Expression $ (4) $ becomes



                $$mathrm{(5)} qquad -sqrt{2w w^2} $$



                Simplify to get the desired result



                $$ -w sqrt{2w} $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 at 18:34









                Marvin CohenMarvin Cohen

                188117




                188117






























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                    Dieringhausen

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                    Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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