When I complete the square on $3x^2 - 12x + 14$ I get an imaginary number, where have I gone wrong?












0












$begingroup$


I have a question in my excersise book:




By completing the square show that the expression $3x^2 - 12x + 14$ is positive for all $x$




My approach was to complete the square and rearrange to make $x$ the subject.



The answer I came to after completing the square was $(sqrt {3}x - 2sqrt{3})^2+2$.



However I get a negative square root:



$$(sqrt {3}x - 2sqrt{3})^2+2 = 0$$
$$(sqrt {3}x - 2sqrt{3})^2 = -2$$
$$sqrt {3}x - 2sqrt{3} = sqrt{-2}$$
$$sqrt{3}x = 2sqrt{3} +- sqrt {-2}$$
$$x = (2sqrt{3} +- sqrt {-2})/3$$



Bad formatting: $+-$ means either $+$ or $-$



Where have I gone wrong?






quadratics square-numbers cubic-equations







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I have a question in my excersise book:




    By completing the square show that the expression $3x^2 - 12x + 14$ is positive for all $x$




    My approach was to complete the square and rearrange to make $x$ the subject.



    The answer I came to after completing the square was $(sqrt {3}x - 2sqrt{3})^2+2$.



    However I get a negative square root:



    $$(sqrt {3}x - 2sqrt{3})^2+2 = 0$$
    $$(sqrt {3}x - 2sqrt{3})^2 = -2$$
    $$sqrt {3}x - 2sqrt{3} = sqrt{-2}$$
    $$sqrt{3}x = 2sqrt{3} +- sqrt {-2}$$
    $$x = (2sqrt{3} +- sqrt {-2})/3$$



    Bad formatting: $+-$ means either $+$ or $-$



    Where have I gone wrong?






    quadratics square-numbers cubic-equations







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I have a question in my excersise book:




      By completing the square show that the expression $3x^2 - 12x + 14$ is positive for all $x$




      My approach was to complete the square and rearrange to make $x$ the subject.



      The answer I came to after completing the square was $(sqrt {3}x - 2sqrt{3})^2+2$.



      However I get a negative square root:



      $$(sqrt {3}x - 2sqrt{3})^2+2 = 0$$
      $$(sqrt {3}x - 2sqrt{3})^2 = -2$$
      $$sqrt {3}x - 2sqrt{3} = sqrt{-2}$$
      $$sqrt{3}x = 2sqrt{3} +- sqrt {-2}$$
      $$x = (2sqrt{3} +- sqrt {-2})/3$$



      Bad formatting: $+-$ means either $+$ or $-$



      Where have I gone wrong?






      quadratics square-numbers cubic-equations







      share|cite|improve this question









      $endgroup$




      I have a question in my excersise book:




      By completing the square show that the expression $3x^2 - 12x + 14$ is positive for all $x$




      My approach was to complete the square and rearrange to make $x$ the subject.



      The answer I came to after completing the square was $(sqrt {3}x - 2sqrt{3})^2+2$.



      However I get a negative square root:



      $$(sqrt {3}x - 2sqrt{3})^2+2 = 0$$
      $$(sqrt {3}x - 2sqrt{3})^2 = -2$$
      $$sqrt {3}x - 2sqrt{3} = sqrt{-2}$$
      $$sqrt{3}x = 2sqrt{3} +- sqrt {-2}$$
      $$x = (2sqrt{3} +- sqrt {-2})/3$$



      Bad formatting: $+-$ means either $+$ or $-$



      Where have I gone wrong?






      quadratics square-numbers cubic-equations




      quadratics square-numbers cubic-equations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 3 '18 at 12:10









      SimonSimon

      1033




      1033






















          4 Answers
          4






          active

          oldest

          votes


















          3












          $begingroup$

          You haven't gone wrong per se. You've just gone a step too far. No need to solve the equation or factor anything. Just note that when you have
          $$
          (sqrt 3x - 2sqrt3)^2 + 2
          $$
          then that's a square (which is non-negative) plus $2$, which necessarily makes the value of the entire expression strictly positive, no matter what $x$ is.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you. That was much easier than I expected it to be, I'll accept your answer as soon as I can
            $endgroup$
            – Simon
            Dec 3 '18 at 12:21



















          3












          $begingroup$

          You haven't gone wrong. By finding the expression for $x$ after completing the square, you are looking for solutions to the equation $3x^2 - 12x + 14 = 0$. You find that there are no (real) solutions, which means that the graph of this parabola never touches the $x$-axis. Because this is a "valley parabola", certainly it will be positive somewhere; that means it is always positive, because it never crosses or touches the $x$-axis.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Do you have to do it exactly that way? Another is to take the factor of 3 (the coefficient of $x^2$) outside, & put it back at the end, to get $$3(x-2)^2+2 .$$ Certainly, this shows just as well that the expression is always positive.






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              In general, remember that a negative radicand does not imply you went wrong anywhere. If anything, it is simply another way to point out that the graph never crosses the $x$-axis (hence no real roots), and since $a > 0$, the graph lies entirely above the $x$-axis, which means the quadratic is always positive.






              share|cite|improve this answer









              $endgroup$













                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%2f3023969%2fwhen-i-complete-the-square-on-3x2-12x-14-i-get-an-imaginary-number-where%23new-answer', 'question_page');
                }
                );

                Post as a guest















                Required, but never shown

























                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                3












                $begingroup$

                You haven't gone wrong per se. You've just gone a step too far. No need to solve the equation or factor anything. Just note that when you have
                $$
                (sqrt 3x - 2sqrt3)^2 + 2
                $$
                then that's a square (which is non-negative) plus $2$, which necessarily makes the value of the entire expression strictly positive, no matter what $x$ is.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Thank you. That was much easier than I expected it to be, I'll accept your answer as soon as I can
                  $endgroup$
                  – Simon
                  Dec 3 '18 at 12:21
















                3












                $begingroup$

                You haven't gone wrong per se. You've just gone a step too far. No need to solve the equation or factor anything. Just note that when you have
                $$
                (sqrt 3x - 2sqrt3)^2 + 2
                $$
                then that's a square (which is non-negative) plus $2$, which necessarily makes the value of the entire expression strictly positive, no matter what $x$ is.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Thank you. That was much easier than I expected it to be, I'll accept your answer as soon as I can
                  $endgroup$
                  – Simon
                  Dec 3 '18 at 12:21














                3












                3








                3





                $begingroup$

                You haven't gone wrong per se. You've just gone a step too far. No need to solve the equation or factor anything. Just note that when you have
                $$
                (sqrt 3x - 2sqrt3)^2 + 2
                $$
                then that's a square (which is non-negative) plus $2$, which necessarily makes the value of the entire expression strictly positive, no matter what $x$ is.






                share|cite|improve this answer









                $endgroup$



                You haven't gone wrong per se. You've just gone a step too far. No need to solve the equation or factor anything. Just note that when you have
                $$
                (sqrt 3x - 2sqrt3)^2 + 2
                $$
                then that's a square (which is non-negative) plus $2$, which necessarily makes the value of the entire expression strictly positive, no matter what $x$ is.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 3 '18 at 12:17









                ArthurArthur

                112k7108192




                112k7108192












                • $begingroup$
                  Thank you. That was much easier than I expected it to be, I'll accept your answer as soon as I can
                  $endgroup$
                  – Simon
                  Dec 3 '18 at 12:21


















                • $begingroup$
                  Thank you. That was much easier than I expected it to be, I'll accept your answer as soon as I can
                  $endgroup$
                  – Simon
                  Dec 3 '18 at 12:21
















                $begingroup$
                Thank you. That was much easier than I expected it to be, I'll accept your answer as soon as I can
                $endgroup$
                – Simon
                Dec 3 '18 at 12:21




                $begingroup$
                Thank you. That was much easier than I expected it to be, I'll accept your answer as soon as I can
                $endgroup$
                – Simon
                Dec 3 '18 at 12:21











                3












                $begingroup$

                You haven't gone wrong. By finding the expression for $x$ after completing the square, you are looking for solutions to the equation $3x^2 - 12x + 14 = 0$. You find that there are no (real) solutions, which means that the graph of this parabola never touches the $x$-axis. Because this is a "valley parabola", certainly it will be positive somewhere; that means it is always positive, because it never crosses or touches the $x$-axis.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  You haven't gone wrong. By finding the expression for $x$ after completing the square, you are looking for solutions to the equation $3x^2 - 12x + 14 = 0$. You find that there are no (real) solutions, which means that the graph of this parabola never touches the $x$-axis. Because this is a "valley parabola", certainly it will be positive somewhere; that means it is always positive, because it never crosses or touches the $x$-axis.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    You haven't gone wrong. By finding the expression for $x$ after completing the square, you are looking for solutions to the equation $3x^2 - 12x + 14 = 0$. You find that there are no (real) solutions, which means that the graph of this parabola never touches the $x$-axis. Because this is a "valley parabola", certainly it will be positive somewhere; that means it is always positive, because it never crosses or touches the $x$-axis.






                    share|cite|improve this answer









                    $endgroup$



                    You haven't gone wrong. By finding the expression for $x$ after completing the square, you are looking for solutions to the equation $3x^2 - 12x + 14 = 0$. You find that there are no (real) solutions, which means that the graph of this parabola never touches the $x$-axis. Because this is a "valley parabola", certainly it will be positive somewhere; that means it is always positive, because it never crosses or touches the $x$-axis.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 3 '18 at 12:20









                    Mees de VriesMees de Vries

                    16.5k12654




                    16.5k12654























                        2












                        $begingroup$

                        Do you have to do it exactly that way? Another is to take the factor of 3 (the coefficient of $x^2$) outside, & put it back at the end, to get $$3(x-2)^2+2 .$$ Certainly, this shows just as well that the expression is always positive.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          Do you have to do it exactly that way? Another is to take the factor of 3 (the coefficient of $x^2$) outside, & put it back at the end, to get $$3(x-2)^2+2 .$$ Certainly, this shows just as well that the expression is always positive.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            Do you have to do it exactly that way? Another is to take the factor of 3 (the coefficient of $x^2$) outside, & put it back at the end, to get $$3(x-2)^2+2 .$$ Certainly, this shows just as well that the expression is always positive.






                            share|cite|improve this answer









                            $endgroup$



                            Do you have to do it exactly that way? Another is to take the factor of 3 (the coefficient of $x^2$) outside, & put it back at the end, to get $$3(x-2)^2+2 .$$ Certainly, this shows just as well that the expression is always positive.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 3 '18 at 12:38









                            AmbretteOrriseyAmbretteOrrisey

                            54210




                            54210























                                1












                                $begingroup$

                                In general, remember that a negative radicand does not imply you went wrong anywhere. If anything, it is simply another way to point out that the graph never crosses the $x$-axis (hence no real roots), and since $a > 0$, the graph lies entirely above the $x$-axis, which means the quadratic is always positive.






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  In general, remember that a negative radicand does not imply you went wrong anywhere. If anything, it is simply another way to point out that the graph never crosses the $x$-axis (hence no real roots), and since $a > 0$, the graph lies entirely above the $x$-axis, which means the quadratic is always positive.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    In general, remember that a negative radicand does not imply you went wrong anywhere. If anything, it is simply another way to point out that the graph never crosses the $x$-axis (hence no real roots), and since $a > 0$, the graph lies entirely above the $x$-axis, which means the quadratic is always positive.






                                    share|cite|improve this answer









                                    $endgroup$



                                    In general, remember that a negative radicand does not imply you went wrong anywhere. If anything, it is simply another way to point out that the graph never crosses the $x$-axis (hence no real roots), and since $a > 0$, the graph lies entirely above the $x$-axis, which means the quadratic is always positive.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 3 '18 at 12:25









                                    KM101KM101

                                    5,8681523




                                    5,8681523






























                                        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%2f3023969%2fwhen-i-complete-the-square-on-3x2-12x-14-i-get-an-imaginary-number-where%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

                                        Bundesstraße 106

                                        Image
                                        Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
                                        Read more

                                        Verónica Boquete

                                        Image
                                        Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
                                        Read more

                                        Ida-Boy-Ed-Garten

                                        Image
                                        Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                                        Read more