Complex quadratic equation always comes out as wrong












1












$begingroup$


So for some reason I always get the wrong answer and I don't understand why



$4z^2-12z+19=0$



I got $z= frac{12 ± sqrt{260}i}{8} $ and the answer is supposed to be $z=frac{3 ± sqrt{10}i}{2} $ where is my mistake?






complex-numbers quadratics







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  • 1




    $begingroup$
    They are the same thing, only a factor of $4$ was cancelled.
    $endgroup$
    – Matti P.
    2 hours ago










  • $begingroup$
    Note that $ sqrt{260} $ (is that is what you meant) is typeset as sqrt{260} .
    $endgroup$
    – Martin R
    2 hours ago
















1












$begingroup$


So for some reason I always get the wrong answer and I don't understand why



$4z^2-12z+19=0$



I got $z= frac{12 ± sqrt{260}i}{8} $ and the answer is supposed to be $z=frac{3 ± sqrt{10}i}{2} $ where is my mistake?






complex-numbers quadratics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    They are the same thing, only a factor of $4$ was cancelled.
    $endgroup$
    – Matti P.
    2 hours ago










  • $begingroup$
    Note that $ sqrt{260} $ (is that is what you meant) is typeset as sqrt{260} .
    $endgroup$
    – Martin R
    2 hours ago














1












1








1





$begingroup$


So for some reason I always get the wrong answer and I don't understand why



$4z^2-12z+19=0$



I got $z= frac{12 ± sqrt{260}i}{8} $ and the answer is supposed to be $z=frac{3 ± sqrt{10}i}{2} $ where is my mistake?






complex-numbers quadratics







share|cite|improve this question











$endgroup$




So for some reason I always get the wrong answer and I don't understand why



$4z^2-12z+19=0$



I got $z= frac{12 ± sqrt{260}i}{8} $ and the answer is supposed to be $z=frac{3 ± sqrt{10}i}{2} $ where is my mistake?






complex-numbers quadratics




complex-numbers quadratics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 12 mins ago









Jonathan Chiang

684




684










asked 2 hours ago









ythhtrgythhtrg

163




163








  • 1




    $begingroup$
    They are the same thing, only a factor of $4$ was cancelled.
    $endgroup$
    – Matti P.
    2 hours ago










  • $begingroup$
    Note that $ sqrt{260} $ (is that is what you meant) is typeset as sqrt{260} .
    $endgroup$
    – Martin R
    2 hours ago














  • 1




    $begingroup$
    They are the same thing, only a factor of $4$ was cancelled.
    $endgroup$
    – Matti P.
    2 hours ago










  • $begingroup$
    Note that $ sqrt{260} $ (is that is what you meant) is typeset as sqrt{260} .
    $endgroup$
    – Martin R
    2 hours ago








1




1




$begingroup$
They are the same thing, only a factor of $4$ was cancelled.
$endgroup$
– Matti P.
2 hours ago




$begingroup$
They are the same thing, only a factor of $4$ was cancelled.
$endgroup$
– Matti P.
2 hours ago












$begingroup$
Note that $ sqrt{260} $ (is that is what you meant) is typeset as sqrt{260} .
$endgroup$
– Martin R
2 hours ago




$begingroup$
Note that $ sqrt{260} $ (is that is what you meant) is typeset as sqrt{260} .
$endgroup$
– Martin R
2 hours ago










3 Answers
3






active

oldest

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4












$begingroup$

Your mistake is that the discriminant should be $12^2 - 4*4*19 = -160$, not $-260$, and then dividing top and bottom by 4 gives the correct answer.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    When the $x$ coefficient is even, say $ax^2+2bx+c=0,$ there is an equivalent version of the quadratic equation which automatically cancels that extra factor of $4.$



    Instead of the usual discriminant $D=b^2-4ac$ with this setup one calculates what might be called the "other" discriminant (for want of a better word) $E=b^2-ac$ as the thing under the radical. As usual if that's negative the complex number $i$ is used when squareroot of discriminant is taken.



    Then the roots are $$frac{-b pm sqrt{E}}{a}.$$



    For the re-written example you have: $4z^2-2(-6)z+19=0,$ we get $E=(-6)^2-4cdot 19=36-4cdot 19=36-76=-40.$ Then roots are $(6 pm sqrt{-40})/12.$ This simplifies again in this case.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      It is $$z_{1,2}=frac{3}{2}pmfrac{sqrt{10}}{2}i$$ by the quadratic formula.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        Your mistake is that the discriminant should be $12^2 - 4*4*19 = -160$, not $-260$, and then dividing top and bottom by 4 gives the correct answer.






        share|cite|improve this answer









        $endgroup$


















          4












          $begingroup$

          Your mistake is that the discriminant should be $12^2 - 4*4*19 = -160$, not $-260$, and then dividing top and bottom by 4 gives the correct answer.






          share|cite|improve this answer









          $endgroup$
















            4












            4








            4





            $begingroup$

            Your mistake is that the discriminant should be $12^2 - 4*4*19 = -160$, not $-260$, and then dividing top and bottom by 4 gives the correct answer.






            share|cite|improve this answer









            $endgroup$



            Your mistake is that the discriminant should be $12^2 - 4*4*19 = -160$, not $-260$, and then dividing top and bottom by 4 gives the correct answer.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 hours ago









            J. W. TannerJ. W. Tanner

            1667




            1667























                2












                $begingroup$

                When the $x$ coefficient is even, say $ax^2+2bx+c=0,$ there is an equivalent version of the quadratic equation which automatically cancels that extra factor of $4.$



                Instead of the usual discriminant $D=b^2-4ac$ with this setup one calculates what might be called the "other" discriminant (for want of a better word) $E=b^2-ac$ as the thing under the radical. As usual if that's negative the complex number $i$ is used when squareroot of discriminant is taken.



                Then the roots are $$frac{-b pm sqrt{E}}{a}.$$



                For the re-written example you have: $4z^2-2(-6)z+19=0,$ we get $E=(-6)^2-4cdot 19=36-4cdot 19=36-76=-40.$ Then roots are $(6 pm sqrt{-40})/12.$ This simplifies again in this case.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  When the $x$ coefficient is even, say $ax^2+2bx+c=0,$ there is an equivalent version of the quadratic equation which automatically cancels that extra factor of $4.$



                  Instead of the usual discriminant $D=b^2-4ac$ with this setup one calculates what might be called the "other" discriminant (for want of a better word) $E=b^2-ac$ as the thing under the radical. As usual if that's negative the complex number $i$ is used when squareroot of discriminant is taken.



                  Then the roots are $$frac{-b pm sqrt{E}}{a}.$$



                  For the re-written example you have: $4z^2-2(-6)z+19=0,$ we get $E=(-6)^2-4cdot 19=36-4cdot 19=36-76=-40.$ Then roots are $(6 pm sqrt{-40})/12.$ This simplifies again in this case.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    When the $x$ coefficient is even, say $ax^2+2bx+c=0,$ there is an equivalent version of the quadratic equation which automatically cancels that extra factor of $4.$



                    Instead of the usual discriminant $D=b^2-4ac$ with this setup one calculates what might be called the "other" discriminant (for want of a better word) $E=b^2-ac$ as the thing under the radical. As usual if that's negative the complex number $i$ is used when squareroot of discriminant is taken.



                    Then the roots are $$frac{-b pm sqrt{E}}{a}.$$



                    For the re-written example you have: $4z^2-2(-6)z+19=0,$ we get $E=(-6)^2-4cdot 19=36-4cdot 19=36-76=-40.$ Then roots are $(6 pm sqrt{-40})/12.$ This simplifies again in this case.






                    share|cite|improve this answer









                    $endgroup$



                    When the $x$ coefficient is even, say $ax^2+2bx+c=0,$ there is an equivalent version of the quadratic equation which automatically cancels that extra factor of $4.$



                    Instead of the usual discriminant $D=b^2-4ac$ with this setup one calculates what might be called the "other" discriminant (for want of a better word) $E=b^2-ac$ as the thing under the radical. As usual if that's negative the complex number $i$ is used when squareroot of discriminant is taken.



                    Then the roots are $$frac{-b pm sqrt{E}}{a}.$$



                    For the re-written example you have: $4z^2-2(-6)z+19=0,$ we get $E=(-6)^2-4cdot 19=36-4cdot 19=36-76=-40.$ Then roots are $(6 pm sqrt{-40})/12.$ This simplifies again in this case.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 hours ago









                    coffeemathcoffeemath

                    2,6781413




                    2,6781413























                        0












                        $begingroup$

                        It is $$z_{1,2}=frac{3}{2}pmfrac{sqrt{10}}{2}i$$ by the quadratic formula.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          It is $$z_{1,2}=frac{3}{2}pmfrac{sqrt{10}}{2}i$$ by the quadratic formula.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            It is $$z_{1,2}=frac{3}{2}pmfrac{sqrt{10}}{2}i$$ by the quadratic formula.






                            share|cite|improve this answer









                            $endgroup$



                            It is $$z_{1,2}=frac{3}{2}pmfrac{sqrt{10}}{2}i$$ by the quadratic formula.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 hours ago









                            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                            73.7k42865




                            73.7k42865






























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