Proving roots to be real

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












Given :



$$(a-b)x^2 + (c-a)x + (a-b) = 0 text where a,b,c in Q $$



How would I prove that the roots of the equation will be real






roots quadratics







share|cite|improve this question





















  • As pointed out by the other answer, it's not always possible. However, if you change the last $(a-b)$ to $(b-a)$, you'd have a much better starting point.
    – Arthur
    Sep 4 at 10:08










  • Use the discriminant.
    – Wuestenfux
    Sep 4 at 10:16














up vote
0
down vote

favorite












Given :



$$(a-b)x^2 + (c-a)x + (a-b) = 0 text where a,b,c in Q $$



How would I prove that the roots of the equation will be real






roots quadratics







share|cite|improve this question





















  • As pointed out by the other answer, it's not always possible. However, if you change the last $(a-b)$ to $(b-a)$, you'd have a much better starting point.
    – Arthur
    Sep 4 at 10:08










  • Use the discriminant.
    – Wuestenfux
    Sep 4 at 10:16












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Given :



$$(a-b)x^2 + (c-a)x + (a-b) = 0 text where a,b,c in Q $$



How would I prove that the roots of the equation will be real






roots quadratics







share|cite|improve this question













Given :



$$(a-b)x^2 + (c-a)x + (a-b) = 0 text where a,b,c in Q $$



How would I prove that the roots of the equation will be real






roots quadratics




roots quadratics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 4 at 10:05









Any3nymous user

101




101











  • As pointed out by the other answer, it's not always possible. However, if you change the last $(a-b)$ to $(b-a)$, you'd have a much better starting point.
    – Arthur
    Sep 4 at 10:08










  • Use the discriminant.
    – Wuestenfux
    Sep 4 at 10:16
















  • As pointed out by the other answer, it's not always possible. However, if you change the last $(a-b)$ to $(b-a)$, you'd have a much better starting point.
    – Arthur
    Sep 4 at 10:08










  • Use the discriminant.
    – Wuestenfux
    Sep 4 at 10:16















As pointed out by the other answer, it's not always possible. However, if you change the last $(a-b)$ to $(b-a)$, you'd have a much better starting point.
– Arthur
Sep 4 at 10:08




As pointed out by the other answer, it's not always possible. However, if you change the last $(a-b)$ to $(b-a)$, you'd have a much better starting point.
– Arthur
Sep 4 at 10:08












Use the discriminant.
– Wuestenfux
Sep 4 at 10:16




Use the discriminant.
– Wuestenfux
Sep 4 at 10:16










2 Answers
2






active

oldest

votes

















up vote
4
down vote













You can't. If $b = 0$ and $a = c = 1$, then the roots of the equation are $pm i$.






share|cite|improve this answer



























    up vote
    0
    down vote













    You have a quadratic equation of the form $ax^2 + bx + c$ being $a,b,c∈Q$ and $a = c$ in quadratic equations have two real solutions if $b^2 - 4ac gt 0$ (Which is the discriminant of the quadratic formula) that's why they are real solutions and not complex solutions. Now change the terms and you'll see.



    Note: You can get a real solution too if $b² - 4ac = 0$, but just one real solution.






    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',
      convertImagesToLinks: true,
      noModals: false,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      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%2f2904860%2fproving-roots-to-be-real%23new-answer', 'question_page');

      );

      Post as a guest

















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      4
      down vote













      You can't. If $b = 0$ and $a = c = 1$, then the roots of the equation are $pm i$.






      share|cite|improve this answer
























        up vote
        4
        down vote













        You can't. If $b = 0$ and $a = c = 1$, then the roots of the equation are $pm i$.






        share|cite|improve this answer






















          up vote
          4
          down vote










          up vote
          4
          down vote









          You can't. If $b = 0$ and $a = c = 1$, then the roots of the equation are $pm i$.






          share|cite|improve this answer












          You can't. If $b = 0$ and $a = c = 1$, then the roots of the equation are $pm i$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 4 at 10:07









          Mees de Vries

          14.3k12348




          14.3k12348




















              up vote
              0
              down vote













              You have a quadratic equation of the form $ax^2 + bx + c$ being $a,b,c∈Q$ and $a = c$ in quadratic equations have two real solutions if $b^2 - 4ac gt 0$ (Which is the discriminant of the quadratic formula) that's why they are real solutions and not complex solutions. Now change the terms and you'll see.



              Note: You can get a real solution too if $b² - 4ac = 0$, but just one real solution.






              share|cite|improve this answer
























                up vote
                0
                down vote













                You have a quadratic equation of the form $ax^2 + bx + c$ being $a,b,c∈Q$ and $a = c$ in quadratic equations have two real solutions if $b^2 - 4ac gt 0$ (Which is the discriminant of the quadratic formula) that's why they are real solutions and not complex solutions. Now change the terms and you'll see.



                Note: You can get a real solution too if $b² - 4ac = 0$, but just one real solution.






                share|cite|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  You have a quadratic equation of the form $ax^2 + bx + c$ being $a,b,c∈Q$ and $a = c$ in quadratic equations have two real solutions if $b^2 - 4ac gt 0$ (Which is the discriminant of the quadratic formula) that's why they are real solutions and not complex solutions. Now change the terms and you'll see.



                  Note: You can get a real solution too if $b² - 4ac = 0$, but just one real solution.






                  share|cite|improve this answer












                  You have a quadratic equation of the form $ax^2 + bx + c$ being $a,b,c∈Q$ and $a = c$ in quadratic equations have two real solutions if $b^2 - 4ac gt 0$ (Which is the discriminant of the quadratic formula) that's why they are real solutions and not complex solutions. Now change the terms and you'll see.



                  Note: You can get a real solution too if $b² - 4ac = 0$, but just one real solution.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 4 at 11:02









                  Enigsis

                  1069




                  1069



























                       

                      draft saved


                      draft discarded















































                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2904860%2fproving-roots-to-be-real%23new-answer', 'question_page');

                      );

                      Post as a guest
































































                      -

                      這個網誌中的熱門文章

                      Is there any way to eliminate the singular point to solve this integral by hand or by approximations?

                      圖片
                      Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite 1 $$int_0^1 (1+(x^-p-1)^1-p)^frac1pdx, pin mathbbR, pge 1$$ I find that $x=0,1$ are singular points(discontinuous points of this integral), is there a way to eliminate those singular points to make sure that this integral can be calculated by hand or by Taylor approximations. In either cases, what are those steps of calculating the integral? calculus real-analysis analysis approximation share | cite | improve this question edited Sep 11 at 5:07 asked Sep 11 at 4:48 Mclalalala 126 8 1 What is $p$? Please give more information to make it complete. – xbh Sep 11 at 5:02 @xbh already fixed it, thanks for reminding. – Mclalalala Sep 11 at 5:10 Seems that $0$ is not singular here… [I might compute wrong] – xbh Sep 11 at 5:30 @xbh can you show me how to get the close formula for the integral above? – Mclalalala Sep 11 at 5:33
                      閱讀完整內容

                      Why am i infinitely getting the same tweet with the Twitter Search API?

                      圖片
                      up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia
                      閱讀完整內容

                      Carbon dioxide

                      圖片
                      "CO2" redirects here. For other uses, see CO2 (disambiguation). Carbon dioxide Names Other names Carbonic acid gas Carbonic anhydride Carbonic oxide Carbon oxide Carbon(IV) oxide Dry ice (solid phase) Identifiers CAS Number 124-38-9   Y 3D model (JSmol) Interactive image Interactive image 3DMet B01131 Beilstein Reference 1900390 ChEBI CHEBI:16526   Y ChEMBL ChEMBL1231871   N ChemSpider 274   Y ECHA InfoCard 100.004.271 EC Number 204-696-9 E number E290 (preservatives) Gmelin Reference 989 KEGG D00004   Y MeSH Carbon+dioxide PubChem CID 280 RTECS number FF6400000 UNII 142M471B3J   Y UN number 1013 (gas), 1845 (solid) InChI InChI=1S/CO2/c2-1-3   Y Key: CURLTUGMZLYLDI-UHFFFAOYSA-N   Y InChI=1/CO2/c2-1-3 Key: CURLTUGMZLYLDI-UHFFFAOYAO SMILES O=C=O C(=O)=O Properties Chemical formula C O 2 Molar mass 44.01 g·mol −1 Appearance Colorless gas Odor Low concentrations: none High concentrations: sharp; acidic [1] Density 1562 kg/m 3 (solid at 1 atm and −78.5 °C) 1101 kg/m 3 (liquid at
                      閱讀完整內容