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












1














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



























    1














    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

























      1












      1








      1







      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













      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

      3177




      3177






















          2 Answers
          2






          active

          oldest

          votes


















          1














          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



















          • 1




            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?
            – Doug Fir
            Jan 6 at 18:00






          • 2




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






          • 1




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



















          1














          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





















            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3064169%2fstruggling-to-simplify-w3-2-sqrt32-w3-2-sqrt50-to-w-sqrt2w%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            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



















            • 1




              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?
              – Doug Fir
              Jan 6 at 18:00






            • 2




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






            • 1




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
















            1














            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



















            • 1




              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?
              – Doug Fir
              Jan 6 at 18:00






            • 2




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






            • 1




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














            1












            1








            1






            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














            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

            2,6211618




            2,6211618










            answered Jan 6 at 17:51









            Thomas ShelbyThomas Shelby

            2,039220




            2,039220








            • 1




              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?
              – Doug Fir
              Jan 6 at 18:00






            • 2




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






            • 1




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














            • 1




              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?
              – Doug Fir
              Jan 6 at 18:00






            • 2




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






            • 1




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








            1




            1




            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?
            – Doug Fir
            Jan 6 at 18:00




            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?
            – Doug Fir
            Jan 6 at 18:00




            2




            2




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




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




            1




            1




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




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











            1














            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


























              1














              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
























                1












                1








                1






                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












                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

                43115




                43115






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3064169%2fstruggling-to-simplify-w3-2-sqrt32-w3-2-sqrt50-to-w-sqrt2w%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Mario Kart Wii

                    Image
                    Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
                    Read more

                    What does “Dominus providebit” mean?

                    Image
                    5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
                    Read more

                    Antonio Litta Visconti Arese

                    Image
                    Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
                    Read more