Quadratics substitute problem












1












$begingroup$


Ok, so I can make the substitution $z^4=u^2$ to solve $az^4+bz^2+c=0,:forall:a,b,cinmathbb{R},:aneq0$?






algebra-precalculus







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  • 2




    $begingroup$
    Yes, you can, together with $z^2=u$.
    $endgroup$
    – José Carlos Santos
    Dec 9 '18 at 22:08










  • $begingroup$
    ok, but then $z^2=|u|$ and I am losing solutions that way!
    $endgroup$
    – Numbers
    Dec 9 '18 at 22:10










  • $begingroup$
    @Numbers have a look at my answer. The two instances of $pm$ indicate there are four solutions, as expected of a fourth degree polynomial. Also you're losing values because instead of taking $z^2=|u|$ you should take $z^2=pm u$
    $endgroup$
    – Rhys Hughes
    Dec 9 '18 at 22:14










  • $begingroup$
    Yeah, u r right, thx man!
    $endgroup$
    – Numbers
    Dec 9 '18 at 22:19
















1












$begingroup$


Ok, so I can make the substitution $z^4=u^2$ to solve $az^4+bz^2+c=0,:forall:a,b,cinmathbb{R},:aneq0$?






algebra-precalculus







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Yes, you can, together with $z^2=u$.
    $endgroup$
    – José Carlos Santos
    Dec 9 '18 at 22:08










  • $begingroup$
    ok, but then $z^2=|u|$ and I am losing solutions that way!
    $endgroup$
    – Numbers
    Dec 9 '18 at 22:10










  • $begingroup$
    @Numbers have a look at my answer. The two instances of $pm$ indicate there are four solutions, as expected of a fourth degree polynomial. Also you're losing values because instead of taking $z^2=|u|$ you should take $z^2=pm u$
    $endgroup$
    – Rhys Hughes
    Dec 9 '18 at 22:14










  • $begingroup$
    Yeah, u r right, thx man!
    $endgroup$
    – Numbers
    Dec 9 '18 at 22:19














1












1








1





$begingroup$


Ok, so I can make the substitution $z^4=u^2$ to solve $az^4+bz^2+c=0,:forall:a,b,cinmathbb{R},:aneq0$?






algebra-precalculus







share|cite|improve this question









$endgroup$




Ok, so I can make the substitution $z^4=u^2$ to solve $az^4+bz^2+c=0,:forall:a,b,cinmathbb{R},:aneq0$?






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 Dec 9 '18 at 22:06









NumbersNumbers

1116




1116








  • 2




    $begingroup$
    Yes, you can, together with $z^2=u$.
    $endgroup$
    – José Carlos Santos
    Dec 9 '18 at 22:08










  • $begingroup$
    ok, but then $z^2=|u|$ and I am losing solutions that way!
    $endgroup$
    – Numbers
    Dec 9 '18 at 22:10










  • $begingroup$
    @Numbers have a look at my answer. The two instances of $pm$ indicate there are four solutions, as expected of a fourth degree polynomial. Also you're losing values because instead of taking $z^2=|u|$ you should take $z^2=pm u$
    $endgroup$
    – Rhys Hughes
    Dec 9 '18 at 22:14










  • $begingroup$
    Yeah, u r right, thx man!
    $endgroup$
    – Numbers
    Dec 9 '18 at 22:19














  • 2




    $begingroup$
    Yes, you can, together with $z^2=u$.
    $endgroup$
    – José Carlos Santos
    Dec 9 '18 at 22:08










  • $begingroup$
    ok, but then $z^2=|u|$ and I am losing solutions that way!
    $endgroup$
    – Numbers
    Dec 9 '18 at 22:10










  • $begingroup$
    @Numbers have a look at my answer. The two instances of $pm$ indicate there are four solutions, as expected of a fourth degree polynomial. Also you're losing values because instead of taking $z^2=|u|$ you should take $z^2=pm u$
    $endgroup$
    – Rhys Hughes
    Dec 9 '18 at 22:14










  • $begingroup$
    Yeah, u r right, thx man!
    $endgroup$
    – Numbers
    Dec 9 '18 at 22:19








2




2




$begingroup$
Yes, you can, together with $z^2=u$.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:08




$begingroup$
Yes, you can, together with $z^2=u$.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:08












$begingroup$
ok, but then $z^2=|u|$ and I am losing solutions that way!
$endgroup$
– Numbers
Dec 9 '18 at 22:10




$begingroup$
ok, but then $z^2=|u|$ and I am losing solutions that way!
$endgroup$
– Numbers
Dec 9 '18 at 22:10












$begingroup$
@Numbers have a look at my answer. The two instances of $pm$ indicate there are four solutions, as expected of a fourth degree polynomial. Also you're losing values because instead of taking $z^2=|u|$ you should take $z^2=pm u$
$endgroup$
– Rhys Hughes
Dec 9 '18 at 22:14




$begingroup$
@Numbers have a look at my answer. The two instances of $pm$ indicate there are four solutions, as expected of a fourth degree polynomial. Also you're losing values because instead of taking $z^2=|u|$ you should take $z^2=pm u$
$endgroup$
– Rhys Hughes
Dec 9 '18 at 22:14












$begingroup$
Yeah, u r right, thx man!
$endgroup$
– Numbers
Dec 9 '18 at 22:19




$begingroup$
Yeah, u r right, thx man!
$endgroup$
– Numbers
Dec 9 '18 at 22:19










1 Answer
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$begingroup$

This is what is known as a biquadratic. Notice we can rewrite it as:



$$a(z^2)^2+b(z^2)^1+c(z^2)^0=0$$ and we have a quadratic. So we use the quadratic equation for:
$$z^2=frac{-bpmsqrt{b^2-4ac}}{2a}to z=pmsqrt{frac{-bpmsqrt{b^2-4ac}}{2a}}$$



In other words, yes, what you are saying is correct.






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    $begingroup$

    This is what is known as a biquadratic. Notice we can rewrite it as:



    $$a(z^2)^2+b(z^2)^1+c(z^2)^0=0$$ and we have a quadratic. So we use the quadratic equation for:
    $$z^2=frac{-bpmsqrt{b^2-4ac}}{2a}to z=pmsqrt{frac{-bpmsqrt{b^2-4ac}}{2a}}$$



    In other words, yes, what you are saying is correct.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      This is what is known as a biquadratic. Notice we can rewrite it as:



      $$a(z^2)^2+b(z^2)^1+c(z^2)^0=0$$ and we have a quadratic. So we use the quadratic equation for:
      $$z^2=frac{-bpmsqrt{b^2-4ac}}{2a}to z=pmsqrt{frac{-bpmsqrt{b^2-4ac}}{2a}}$$



      In other words, yes, what you are saying is correct.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        This is what is known as a biquadratic. Notice we can rewrite it as:



        $$a(z^2)^2+b(z^2)^1+c(z^2)^0=0$$ and we have a quadratic. So we use the quadratic equation for:
        $$z^2=frac{-bpmsqrt{b^2-4ac}}{2a}to z=pmsqrt{frac{-bpmsqrt{b^2-4ac}}{2a}}$$



        In other words, yes, what you are saying is correct.






        share|cite|improve this answer









        $endgroup$



        This is what is known as a biquadratic. Notice we can rewrite it as:



        $$a(z^2)^2+b(z^2)^1+c(z^2)^0=0$$ and we have a quadratic. So we use the quadratic equation for:
        $$z^2=frac{-bpmsqrt{b^2-4ac}}{2a}to z=pmsqrt{frac{-bpmsqrt{b^2-4ac}}{2a}}$$



        In other words, yes, what you are saying is correct.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 22:10









        Rhys HughesRhys Hughes

        6,7001530




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