Complex Solutions to a Quadratic Equation












1












$begingroup$



Given equation:
$$x^2 + 5x + c = 0$$
For what $c$'s will the equation have non-real solutions?




I have try to plug in all kinds of different values for $c$ and then solve the equation to see if it gives me complex solution.



My question: Is there a theorem/lemma or a formula I need to state to answer this question?






algebra-precalculus







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$endgroup$












  • $begingroup$
    Hint: Just use the Quadratic Formula and analyze that result for the roots.
    $endgroup$
    – Moo
    Sep 16 '16 at 1:26








  • 1




    $begingroup$
    Try plotting it for $c=0$, and consider what happens to the roots if $c$ increases or decreases.
    $endgroup$
    – Semiclassical
    Sep 16 '16 at 1:29
















1












$begingroup$



Given equation:
$$x^2 + 5x + c = 0$$
For what $c$'s will the equation have non-real solutions?




I have try to plug in all kinds of different values for $c$ and then solve the equation to see if it gives me complex solution.



My question: Is there a theorem/lemma or a formula I need to state to answer this question?






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: Just use the Quadratic Formula and analyze that result for the roots.
    $endgroup$
    – Moo
    Sep 16 '16 at 1:26








  • 1




    $begingroup$
    Try plotting it for $c=0$, and consider what happens to the roots if $c$ increases or decreases.
    $endgroup$
    – Semiclassical
    Sep 16 '16 at 1:29














1












1








1





$begingroup$



Given equation:
$$x^2 + 5x + c = 0$$
For what $c$'s will the equation have non-real solutions?




I have try to plug in all kinds of different values for $c$ and then solve the equation to see if it gives me complex solution.



My question: Is there a theorem/lemma or a formula I need to state to answer this question?






algebra-precalculus







share|cite|improve this question











$endgroup$





Given equation:
$$x^2 + 5x + c = 0$$
For what $c$'s will the equation have non-real solutions?




I have try to plug in all kinds of different values for $c$ and then solve the equation to see if it gives me complex solution.



My question: Is there a theorem/lemma or a formula I need to state to answer this question?






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 5:37









Za Ira

161115




161115










asked Sep 16 '16 at 1:25









Cesar AgamaCesar Agama

5414




5414












  • $begingroup$
    Hint: Just use the Quadratic Formula and analyze that result for the roots.
    $endgroup$
    – Moo
    Sep 16 '16 at 1:26








  • 1




    $begingroup$
    Try plotting it for $c=0$, and consider what happens to the roots if $c$ increases or decreases.
    $endgroup$
    – Semiclassical
    Sep 16 '16 at 1:29


















  • $begingroup$
    Hint: Just use the Quadratic Formula and analyze that result for the roots.
    $endgroup$
    – Moo
    Sep 16 '16 at 1:26








  • 1




    $begingroup$
    Try plotting it for $c=0$, and consider what happens to the roots if $c$ increases or decreases.
    $endgroup$
    – Semiclassical
    Sep 16 '16 at 1:29
















$begingroup$
Hint: Just use the Quadratic Formula and analyze that result for the roots.
$endgroup$
– Moo
Sep 16 '16 at 1:26






$begingroup$
Hint: Just use the Quadratic Formula and analyze that result for the roots.
$endgroup$
– Moo
Sep 16 '16 at 1:26






1




1




$begingroup$
Try plotting it for $c=0$, and consider what happens to the roots if $c$ increases or decreases.
$endgroup$
– Semiclassical
Sep 16 '16 at 1:29




$begingroup$
Try plotting it for $c=0$, and consider what happens to the roots if $c$ increases or decreases.
$endgroup$
– Semiclassical
Sep 16 '16 at 1:29










3 Answers
3






active

oldest

votes


















3












$begingroup$

You have exactly one root if you have a perfect square.



Compete the square:



$(x^2 + 2 frac 52 x + frac {25}{4}) = (x+frac 52)^2$



If $c = frac {25}{4}$ there is one (real) root.



If $c >frac {25}{4}$ your curve never touches the x axis. And you have no real roots. If a quadratic has no real roots, it has complex roots.



if $c > frac {25}{4}$ the curve crosses the x axis. And it crosses on both sides of the vertex.



Alternate: apply the quadratic formula



If you know the quadratic formula, then what does it take to get a negative under the radical?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Great explanation. I was not thinking about square root or graphing it. Thanks!
    $endgroup$
    – Cesar Agama
    Sep 16 '16 at 6:19



















2












$begingroup$

Hint: This question is equivalent to finding the values of $c$ when $$b^2-4ac=5^2-4c<0$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    The discriminant formula for any quadratic is-
    $D=b^2-4ac$, where $a,b$ and $c$ are the respective coefficients of the quadratic.




    • If $D>0$, then the quadratic has two real and distinct roots.

    • If $D=0$, then the quadratic has two real and equal roots.

    • If $D<0$, then the quadratic has two complex roots.



    In your question-
    $$a=1, b=5, c=?$$
    $$D=5^2 - 4×1×c =25 - 4c$$



    The equation has non-real roots only if-
    $D<0$
    $Rightarrow 25 - 4c <0$
    $Rightarrow c < 25/4$ $_{(Answer)}$






    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









      3












      $begingroup$

      You have exactly one root if you have a perfect square.



      Compete the square:



      $(x^2 + 2 frac 52 x + frac {25}{4}) = (x+frac 52)^2$



      If $c = frac {25}{4}$ there is one (real) root.



      If $c >frac {25}{4}$ your curve never touches the x axis. And you have no real roots. If a quadratic has no real roots, it has complex roots.



      if $c > frac {25}{4}$ the curve crosses the x axis. And it crosses on both sides of the vertex.



      Alternate: apply the quadratic formula



      If you know the quadratic formula, then what does it take to get a negative under the radical?






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Great explanation. I was not thinking about square root or graphing it. Thanks!
        $endgroup$
        – Cesar Agama
        Sep 16 '16 at 6:19
















      3












      $begingroup$

      You have exactly one root if you have a perfect square.



      Compete the square:



      $(x^2 + 2 frac 52 x + frac {25}{4}) = (x+frac 52)^2$



      If $c = frac {25}{4}$ there is one (real) root.



      If $c >frac {25}{4}$ your curve never touches the x axis. And you have no real roots. If a quadratic has no real roots, it has complex roots.



      if $c > frac {25}{4}$ the curve crosses the x axis. And it crosses on both sides of the vertex.



      Alternate: apply the quadratic formula



      If you know the quadratic formula, then what does it take to get a negative under the radical?






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Great explanation. I was not thinking about square root or graphing it. Thanks!
        $endgroup$
        – Cesar Agama
        Sep 16 '16 at 6:19














      3












      3








      3





      $begingroup$

      You have exactly one root if you have a perfect square.



      Compete the square:



      $(x^2 + 2 frac 52 x + frac {25}{4}) = (x+frac 52)^2$



      If $c = frac {25}{4}$ there is one (real) root.



      If $c >frac {25}{4}$ your curve never touches the x axis. And you have no real roots. If a quadratic has no real roots, it has complex roots.



      if $c > frac {25}{4}$ the curve crosses the x axis. And it crosses on both sides of the vertex.



      Alternate: apply the quadratic formula



      If you know the quadratic formula, then what does it take to get a negative under the radical?






      share|cite|improve this answer









      $endgroup$



      You have exactly one root if you have a perfect square.



      Compete the square:



      $(x^2 + 2 frac 52 x + frac {25}{4}) = (x+frac 52)^2$



      If $c = frac {25}{4}$ there is one (real) root.



      If $c >frac {25}{4}$ your curve never touches the x axis. And you have no real roots. If a quadratic has no real roots, it has complex roots.



      if $c > frac {25}{4}$ the curve crosses the x axis. And it crosses on both sides of the vertex.



      Alternate: apply the quadratic formula



      If you know the quadratic formula, then what does it take to get a negative under the radical?







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Sep 16 '16 at 1:33









      Doug MDoug M

      44.5k31854




      44.5k31854












      • $begingroup$
        Great explanation. I was not thinking about square root or graphing it. Thanks!
        $endgroup$
        – Cesar Agama
        Sep 16 '16 at 6:19


















      • $begingroup$
        Great explanation. I was not thinking about square root or graphing it. Thanks!
        $endgroup$
        – Cesar Agama
        Sep 16 '16 at 6:19
















      $begingroup$
      Great explanation. I was not thinking about square root or graphing it. Thanks!
      $endgroup$
      – Cesar Agama
      Sep 16 '16 at 6:19




      $begingroup$
      Great explanation. I was not thinking about square root or graphing it. Thanks!
      $endgroup$
      – Cesar Agama
      Sep 16 '16 at 6:19











      2












      $begingroup$

      Hint: This question is equivalent to finding the values of $c$ when $$b^2-4ac=5^2-4c<0$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Hint: This question is equivalent to finding the values of $c$ when $$b^2-4ac=5^2-4c<0$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Hint: This question is equivalent to finding the values of $c$ when $$b^2-4ac=5^2-4c<0$$






          share|cite|improve this answer









          $endgroup$



          Hint: This question is equivalent to finding the values of $c$ when $$b^2-4ac=5^2-4c<0$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 16 '16 at 1:32









          Siong Thye GohSiong Thye Goh

          100k1466117




          100k1466117























              1












              $begingroup$

              The discriminant formula for any quadratic is-
              $D=b^2-4ac$, where $a,b$ and $c$ are the respective coefficients of the quadratic.




              • If $D>0$, then the quadratic has two real and distinct roots.

              • If $D=0$, then the quadratic has two real and equal roots.

              • If $D<0$, then the quadratic has two complex roots.



              In your question-
              $$a=1, b=5, c=?$$
              $$D=5^2 - 4×1×c =25 - 4c$$



              The equation has non-real roots only if-
              $D<0$
              $Rightarrow 25 - 4c <0$
              $Rightarrow c < 25/4$ $_{(Answer)}$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                The discriminant formula for any quadratic is-
                $D=b^2-4ac$, where $a,b$ and $c$ are the respective coefficients of the quadratic.




                • If $D>0$, then the quadratic has two real and distinct roots.

                • If $D=0$, then the quadratic has two real and equal roots.

                • If $D<0$, then the quadratic has two complex roots.



                In your question-
                $$a=1, b=5, c=?$$
                $$D=5^2 - 4×1×c =25 - 4c$$



                The equation has non-real roots only if-
                $D<0$
                $Rightarrow 25 - 4c <0$
                $Rightarrow c < 25/4$ $_{(Answer)}$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The discriminant formula for any quadratic is-
                  $D=b^2-4ac$, where $a,b$ and $c$ are the respective coefficients of the quadratic.




                  • If $D>0$, then the quadratic has two real and distinct roots.

                  • If $D=0$, then the quadratic has two real and equal roots.

                  • If $D<0$, then the quadratic has two complex roots.



                  In your question-
                  $$a=1, b=5, c=?$$
                  $$D=5^2 - 4×1×c =25 - 4c$$



                  The equation has non-real roots only if-
                  $D<0$
                  $Rightarrow 25 - 4c <0$
                  $Rightarrow c < 25/4$ $_{(Answer)}$






                  share|cite|improve this answer











                  $endgroup$



                  The discriminant formula for any quadratic is-
                  $D=b^2-4ac$, where $a,b$ and $c$ are the respective coefficients of the quadratic.




                  • If $D>0$, then the quadratic has two real and distinct roots.

                  • If $D=0$, then the quadratic has two real and equal roots.

                  • If $D<0$, then the quadratic has two complex roots.



                  In your question-
                  $$a=1, b=5, c=?$$
                  $$D=5^2 - 4×1×c =25 - 4c$$



                  The equation has non-real roots only if-
                  $D<0$
                  $Rightarrow 25 - 4c <0$
                  $Rightarrow c < 25/4$ $_{(Answer)}$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 6 at 7:48









                  Za Ira

                  161115




                  161115










                  answered Jan 6 at 5:58









                  user629353user629353

                  1187




                  1187






























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