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







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












$begingroup$



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











    $endgroup$













      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%2f1928687%2fcomplex-solutions-to-a-quadratic-equation%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      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






























                      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%2f1928687%2fcomplex-solutions-to-a-quadratic-equation%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

                      Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

                      Image
                      28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
                      Read more

                      ts Property 'filter' does not exist on type '{}'

                      Image
                      1 I'm trying to follow Angular Material guide to create a custom filter for autocomplete input text https://material.angular.io/components/autocomplete/overview This is what I have TS import { Component, OnInit } from '@angular/core'; import { TemplateRef } from '@angular/core'; import { FormControl } from "@angular/forms"; import { Observable } from "rxjs"; import { map, startWith, filter } from 'rxjs/operators'; (...) export class AppComponent implements OnInit { title = 'Your Profile'; displayedColumns: string = ['selected', 'name', 'description', 'pnr']; dataSource = ELEMENT_DATA; myControl = new FormControl(); options: any = ['One', 'Two', 'Three']; filteredOptions:
                      Read more

                      mat-slide-toggle shouldn't change it's state when I click cancel in confirmation window

                      Image
                      1 When I checked the slide toggle it should work as expected. But when I try to uncheck it there is a confirmation window asking if you are sure. When I click cancel in that confirmation window, still the toggle changes to unchecked, which shouldn't happen. It should stay in the same state. Here is my Html and TS Code // html <mat-slide-toggle (change)="change($event)" [checked]="isChecked()" > To-pay </mat-slide-toggle> TS code: // ts change(e) { if(this.checked) { if(confirm("Are you sure")) { console.log("toggle") } else { e.stopPropagation(); console.log("toggle should not change if I click the cancel button") }
                      Read more