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







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  • $begingroup$
    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






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$begingroup$

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









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      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






























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