Relation of Roots of Bi-quadratic equation without using Vieta's formula












1












$begingroup$



If $ a, b, c, d $ are roots of the equation
$$ x^4 + px^3 + qx^2 + rx + s=0 $$
Show that
$$ (1+a^2)(1+b^2)(1+c^2)(1+d^2)=(1+s-q)^2 + (p-r)^2 $$




I was able to show this by using Vieta's formulas but the book from which I am doing this has not yet taught, till now it has just taught reminder, factor theorem and fundamental theorem of algebra so I want to ask that is there any other solution using only these or some more interesting way to do






polynomials complex-numbers roots







share|cite|improve this question











$endgroup$












  • $begingroup$
    See math.stackexchange.com/questions/3063351/…
    $endgroup$
    – lab bhattacharjee
    Jan 6 at 18:12
















1












$begingroup$



If $ a, b, c, d $ are roots of the equation
$$ x^4 + px^3 + qx^2 + rx + s=0 $$
Show that
$$ (1+a^2)(1+b^2)(1+c^2)(1+d^2)=(1+s-q)^2 + (p-r)^2 $$




I was able to show this by using Vieta's formulas but the book from which I am doing this has not yet taught, till now it has just taught reminder, factor theorem and fundamental theorem of algebra so I want to ask that is there any other solution using only these or some more interesting way to do






polynomials complex-numbers roots







share|cite|improve this question











$endgroup$












  • $begingroup$
    See math.stackexchange.com/questions/3063351/…
    $endgroup$
    – lab bhattacharjee
    Jan 6 at 18:12














1












1








1





$begingroup$



If $ a, b, c, d $ are roots of the equation
$$ x^4 + px^3 + qx^2 + rx + s=0 $$
Show that
$$ (1+a^2)(1+b^2)(1+c^2)(1+d^2)=(1+s-q)^2 + (p-r)^2 $$




I was able to show this by using Vieta's formulas but the book from which I am doing this has not yet taught, till now it has just taught reminder, factor theorem and fundamental theorem of algebra so I want to ask that is there any other solution using only these or some more interesting way to do






polynomials complex-numbers roots







share|cite|improve this question











$endgroup$





If $ a, b, c, d $ are roots of the equation
$$ x^4 + px^3 + qx^2 + rx + s=0 $$
Show that
$$ (1+a^2)(1+b^2)(1+c^2)(1+d^2)=(1+s-q)^2 + (p-r)^2 $$




I was able to show this by using Vieta's formulas but the book from which I am doing this has not yet taught, till now it has just taught reminder, factor theorem and fundamental theorem of algebra so I want to ask that is there any other solution using only these or some more interesting way to do






polynomials complex-numbers roots




polynomials complex-numbers roots






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 18:19









greedoid

38.8k114797




38.8k114797










asked Jan 6 at 17:59









Keshav SharmaKeshav Sharma

985




985












  • $begingroup$
    See math.stackexchange.com/questions/3063351/…
    $endgroup$
    – lab bhattacharjee
    Jan 6 at 18:12


















  • $begingroup$
    See math.stackexchange.com/questions/3063351/…
    $endgroup$
    – lab bhattacharjee
    Jan 6 at 18:12
















$begingroup$
See math.stackexchange.com/questions/3063351/…
$endgroup$
– lab bhattacharjee
Jan 6 at 18:12




$begingroup$
See math.stackexchange.com/questions/3063351/…
$endgroup$
– lab bhattacharjee
Jan 6 at 18:12










2 Answers
2






active

oldest

votes


















2












$begingroup$

So: $$ f(x)= (x-a)(x-b)(x-c)(x-d)$$



$$f(i) = (i-a)...;;;;{rm and };;;;f(-i)= (-i-a)...$$
thus $$f(i)f(-i) = (a^2+1)(b^2+1)(c^2+1)(d^2+1)$$



On the other side $$f(i)f(-i) = (1 -ip -q+ ri + s) (1 + ip + q - ri + s)$$
$$ = (1-q+s-i(p-r))(1-q+s+i(p-r))$$ $$= (1-q+s)^2+(p-r)^2$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Also, we can use $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2.$$
    Indeed,
    $$(1+a^2)(1+b^2)(1+c^2)(1+d^2)=((a+b)^2+(1-ab)^2)((c+d)^2+(1-cd)^2)=$$
    $$=((a+b)(c+d)-(1-ab)(1-cd))^2+((a+b)(1-cd)+(c+d)(1-ab))^2=$$
    $$=(ab+ac+ad+bc+bd+cd-abcd-1)^2+(a+b+c+d-abc-abd-acd-bcd)^2=$$
    $$=(q-s-1)^2+(-p+r)^2=(1+s-q)^2+(p-r)^2.$$






    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%2f3064194%2frelation-of-roots-of-bi-quadratic-equation-without-using-vietas-formula%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









      2












      $begingroup$

      So: $$ f(x)= (x-a)(x-b)(x-c)(x-d)$$



      $$f(i) = (i-a)...;;;;{rm and };;;;f(-i)= (-i-a)...$$
      thus $$f(i)f(-i) = (a^2+1)(b^2+1)(c^2+1)(d^2+1)$$



      On the other side $$f(i)f(-i) = (1 -ip -q+ ri + s) (1 + ip + q - ri + s)$$
      $$ = (1-q+s-i(p-r))(1-q+s+i(p-r))$$ $$= (1-q+s)^2+(p-r)^2$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        So: $$ f(x)= (x-a)(x-b)(x-c)(x-d)$$



        $$f(i) = (i-a)...;;;;{rm and };;;;f(-i)= (-i-a)...$$
        thus $$f(i)f(-i) = (a^2+1)(b^2+1)(c^2+1)(d^2+1)$$



        On the other side $$f(i)f(-i) = (1 -ip -q+ ri + s) (1 + ip + q - ri + s)$$
        $$ = (1-q+s-i(p-r))(1-q+s+i(p-r))$$ $$= (1-q+s)^2+(p-r)^2$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          So: $$ f(x)= (x-a)(x-b)(x-c)(x-d)$$



          $$f(i) = (i-a)...;;;;{rm and };;;;f(-i)= (-i-a)...$$
          thus $$f(i)f(-i) = (a^2+1)(b^2+1)(c^2+1)(d^2+1)$$



          On the other side $$f(i)f(-i) = (1 -ip -q+ ri + s) (1 + ip + q - ri + s)$$
          $$ = (1-q+s-i(p-r))(1-q+s+i(p-r))$$ $$= (1-q+s)^2+(p-r)^2$$






          share|cite|improve this answer









          $endgroup$



          So: $$ f(x)= (x-a)(x-b)(x-c)(x-d)$$



          $$f(i) = (i-a)...;;;;{rm and };;;;f(-i)= (-i-a)...$$
          thus $$f(i)f(-i) = (a^2+1)(b^2+1)(c^2+1)(d^2+1)$$



          On the other side $$f(i)f(-i) = (1 -ip -q+ ri + s) (1 + ip + q - ri + s)$$
          $$ = (1-q+s-i(p-r))(1-q+s+i(p-r))$$ $$= (1-q+s)^2+(p-r)^2$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 6 at 18:14









          greedoidgreedoid

          38.8k114797




          38.8k114797























              0












              $begingroup$

              Also, we can use $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2.$$
              Indeed,
              $$(1+a^2)(1+b^2)(1+c^2)(1+d^2)=((a+b)^2+(1-ab)^2)((c+d)^2+(1-cd)^2)=$$
              $$=((a+b)(c+d)-(1-ab)(1-cd))^2+((a+b)(1-cd)+(c+d)(1-ab))^2=$$
              $$=(ab+ac+ad+bc+bd+cd-abcd-1)^2+(a+b+c+d-abc-abd-acd-bcd)^2=$$
              $$=(q-s-1)^2+(-p+r)^2=(1+s-q)^2+(p-r)^2.$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Also, we can use $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2.$$
                Indeed,
                $$(1+a^2)(1+b^2)(1+c^2)(1+d^2)=((a+b)^2+(1-ab)^2)((c+d)^2+(1-cd)^2)=$$
                $$=((a+b)(c+d)-(1-ab)(1-cd))^2+((a+b)(1-cd)+(c+d)(1-ab))^2=$$
                $$=(ab+ac+ad+bc+bd+cd-abcd-1)^2+(a+b+c+d-abc-abd-acd-bcd)^2=$$
                $$=(q-s-1)^2+(-p+r)^2=(1+s-q)^2+(p-r)^2.$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Also, we can use $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2.$$
                  Indeed,
                  $$(1+a^2)(1+b^2)(1+c^2)(1+d^2)=((a+b)^2+(1-ab)^2)((c+d)^2+(1-cd)^2)=$$
                  $$=((a+b)(c+d)-(1-ab)(1-cd))^2+((a+b)(1-cd)+(c+d)(1-ab))^2=$$
                  $$=(ab+ac+ad+bc+bd+cd-abcd-1)^2+(a+b+c+d-abc-abd-acd-bcd)^2=$$
                  $$=(q-s-1)^2+(-p+r)^2=(1+s-q)^2+(p-r)^2.$$






                  share|cite|improve this answer









                  $endgroup$



                  Also, we can use $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2.$$
                  Indeed,
                  $$(1+a^2)(1+b^2)(1+c^2)(1+d^2)=((a+b)^2+(1-ab)^2)((c+d)^2+(1-cd)^2)=$$
                  $$=((a+b)(c+d)-(1-ab)(1-cd))^2+((a+b)(1-cd)+(c+d)(1-ab))^2=$$
                  $$=(ab+ac+ad+bc+bd+cd-abcd-1)^2+(a+b+c+d-abc-abd-acd-bcd)^2=$$
                  $$=(q-s-1)^2+(-p+r)^2=(1+s-q)^2+(p-r)^2.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 6 at 19:21









                  Michael RozenbergMichael Rozenberg

                  98.3k1590188




                  98.3k1590188






























                      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%2f3064194%2frelation-of-roots-of-bi-quadratic-equation-without-using-vietas-formula%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

                      An IMO inspired problem

                      Image
                      14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
                      Read more

                      Management

                      Image
                      Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
                      Read more

                      Has there ever been an instance of an active nuclear power plant within or near a war zone?

                      Image
                      36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
                      Read more