Find The Probability of Quadratic having imaginary roots












1














Two numbers$ p$ and$ q$ are both chosen randomly (and independently of each other) from the interval$ [-2, 2]$. Find the probability that$ 4x^2+4px+1-q^2=0$ has imaginary roots.



How do you solve this problem? Given that we're trying to find out when the quadratic has imaginary roots I suppose we use the discriminant. Which, $ b^2-4ac=(4p)^2-4(4)(1-q^2)$. It can then be factored out to$ 4^2(p^2-(1-q^2)=4^2(p^2-(1-q)(1+q))$



I'm not even sure if I'm on the right track here. The answer given to us was $ frac{pi}{16}$ and I'm still at a lost on how to get there. An explanation would be appreciated.






polynomials







share|cite|improve this question


















  • 1




    Hint: $p^2-(1-q^2) < 0 Rightarrow p^2 + q^2 < 1$
    – gandalf61
    Dec 3 '18 at 15:00










  • As @timtfj points out in an aswer, this should read "non-real roots" - if you really mean imaginary roots (no real part), then the probability is 0.
    – NickD
    Dec 3 '18 at 15:30


















1














Two numbers$ p$ and$ q$ are both chosen randomly (and independently of each other) from the interval$ [-2, 2]$. Find the probability that$ 4x^2+4px+1-q^2=0$ has imaginary roots.



How do you solve this problem? Given that we're trying to find out when the quadratic has imaginary roots I suppose we use the discriminant. Which, $ b^2-4ac=(4p)^2-4(4)(1-q^2)$. It can then be factored out to$ 4^2(p^2-(1-q^2)=4^2(p^2-(1-q)(1+q))$



I'm not even sure if I'm on the right track here. The answer given to us was $ frac{pi}{16}$ and I'm still at a lost on how to get there. An explanation would be appreciated.






polynomials







share|cite|improve this question


















  • 1




    Hint: $p^2-(1-q^2) < 0 Rightarrow p^2 + q^2 < 1$
    – gandalf61
    Dec 3 '18 at 15:00










  • As @timtfj points out in an aswer, this should read "non-real roots" - if you really mean imaginary roots (no real part), then the probability is 0.
    – NickD
    Dec 3 '18 at 15:30
















1












1








1







Two numbers$ p$ and$ q$ are both chosen randomly (and independently of each other) from the interval$ [-2, 2]$. Find the probability that$ 4x^2+4px+1-q^2=0$ has imaginary roots.



How do you solve this problem? Given that we're trying to find out when the quadratic has imaginary roots I suppose we use the discriminant. Which, $ b^2-4ac=(4p)^2-4(4)(1-q^2)$. It can then be factored out to$ 4^2(p^2-(1-q^2)=4^2(p^2-(1-q)(1+q))$



I'm not even sure if I'm on the right track here. The answer given to us was $ frac{pi}{16}$ and I'm still at a lost on how to get there. An explanation would be appreciated.






polynomials







share|cite|improve this question













Two numbers$ p$ and$ q$ are both chosen randomly (and independently of each other) from the interval$ [-2, 2]$. Find the probability that$ 4x^2+4px+1-q^2=0$ has imaginary roots.



How do you solve this problem? Given that we're trying to find out when the quadratic has imaginary roots I suppose we use the discriminant. Which, $ b^2-4ac=(4p)^2-4(4)(1-q^2)$. It can then be factored out to$ 4^2(p^2-(1-q^2)=4^2(p^2-(1-q)(1+q))$



I'm not even sure if I'm on the right track here. The answer given to us was $ frac{pi}{16}$ and I'm still at a lost on how to get there. An explanation would be appreciated.






polynomials




polynomials






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 14:45









SolvingTrainee

174




174








  • 1




    Hint: $p^2-(1-q^2) < 0 Rightarrow p^2 + q^2 < 1$
    – gandalf61
    Dec 3 '18 at 15:00










  • As @timtfj points out in an aswer, this should read "non-real roots" - if you really mean imaginary roots (no real part), then the probability is 0.
    – NickD
    Dec 3 '18 at 15:30
















  • 1




    Hint: $p^2-(1-q^2) < 0 Rightarrow p^2 + q^2 < 1$
    – gandalf61
    Dec 3 '18 at 15:00










  • As @timtfj points out in an aswer, this should read "non-real roots" - if you really mean imaginary roots (no real part), then the probability is 0.
    – NickD
    Dec 3 '18 at 15:30










1




1




Hint: $p^2-(1-q^2) < 0 Rightarrow p^2 + q^2 < 1$
– gandalf61
Dec 3 '18 at 15:00




Hint: $p^2-(1-q^2) < 0 Rightarrow p^2 + q^2 < 1$
– gandalf61
Dec 3 '18 at 15:00












As @timtfj points out in an aswer, this should read "non-real roots" - if you really mean imaginary roots (no real part), then the probability is 0.
– NickD
Dec 3 '18 at 15:30






As @timtfj points out in an aswer, this should read "non-real roots" - if you really mean imaginary roots (no real part), then the probability is 0.
– NickD
Dec 3 '18 at 15:30












4 Answers
4






active

oldest

votes


















3














If there are two imaginary roots to a quadratic equation, then the discriminant is negative.



$$b^2-4ac=16p^2-16+16q^2=16(p^2-1+q^2)<0$$
$$p^2+q^2<1$$



The probability is the ratio between the area inside the circle defined by $p^2+q^2=1$ and the area of the rectangle which holds all possible values of $p,q$. The circle is a circle of radius 1, and therefore has an area of $pi$. The rectangle is a square of side length 4, and therefore has an area of 16. The ratio is therefore $frac{pi}{16}$






share|cite|improve this answer





























    3














    Imagine $p$ and $q$ plotted along horizontal and vertical axes respectively. The given interval for $p$ and $q$ corresponds to a square of area 16. The discriminant is negative when $p^2 + q^2 lt 1$ which corresponds to a circle of area $pi$. The probability of non-real roots is the ratio of the area of the circle to the area of the square.






    share|cite|improve this answer































      2














      To nitpick:



      I think the problem is incorrectly worded and we should be finding the probability of non-real roots: imaginary ones would be complex numbers $a+bi$ with $a=0.$



      For the given equation, this only happens when $p=0$, so unless $p$ is chosen from finitely many values rather than from all the real numbers in $[-2,2]$, the probability is zero if we take imaginary literally.






      share|cite|improve this answer































        0














        Yes, you have to use the discriminant $D(p,q)$ of that polynomial. And you have to prove that if $A$ is the area of the region$$left{(p,q)in[-2,2]times[-2,2],middle|,D(p,q)<0right},$$then$$frac A{16}=fracpi4,$$where that $16$ is the area of the square $[-2,2]times[-2,2]$.






        share|cite|improve this answer





















          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%2f3024149%2ffind-the-probability-of-quadratic-having-imaginary-roots%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3














          If there are two imaginary roots to a quadratic equation, then the discriminant is negative.



          $$b^2-4ac=16p^2-16+16q^2=16(p^2-1+q^2)<0$$
          $$p^2+q^2<1$$



          The probability is the ratio between the area inside the circle defined by $p^2+q^2=1$ and the area of the rectangle which holds all possible values of $p,q$. The circle is a circle of radius 1, and therefore has an area of $pi$. The rectangle is a square of side length 4, and therefore has an area of 16. The ratio is therefore $frac{pi}{16}$






          share|cite|improve this answer


























            3














            If there are two imaginary roots to a quadratic equation, then the discriminant is negative.



            $$b^2-4ac=16p^2-16+16q^2=16(p^2-1+q^2)<0$$
            $$p^2+q^2<1$$



            The probability is the ratio between the area inside the circle defined by $p^2+q^2=1$ and the area of the rectangle which holds all possible values of $p,q$. The circle is a circle of radius 1, and therefore has an area of $pi$. The rectangle is a square of side length 4, and therefore has an area of 16. The ratio is therefore $frac{pi}{16}$






            share|cite|improve this answer
























              3












              3








              3






              If there are two imaginary roots to a quadratic equation, then the discriminant is negative.



              $$b^2-4ac=16p^2-16+16q^2=16(p^2-1+q^2)<0$$
              $$p^2+q^2<1$$



              The probability is the ratio between the area inside the circle defined by $p^2+q^2=1$ and the area of the rectangle which holds all possible values of $p,q$. The circle is a circle of radius 1, and therefore has an area of $pi$. The rectangle is a square of side length 4, and therefore has an area of 16. The ratio is therefore $frac{pi}{16}$






              share|cite|improve this answer












              If there are two imaginary roots to a quadratic equation, then the discriminant is negative.



              $$b^2-4ac=16p^2-16+16q^2=16(p^2-1+q^2)<0$$
              $$p^2+q^2<1$$



              The probability is the ratio between the area inside the circle defined by $p^2+q^2=1$ and the area of the rectangle which holds all possible values of $p,q$. The circle is a circle of radius 1, and therefore has an area of $pi$. The rectangle is a square of side length 4, and therefore has an area of 16. The ratio is therefore $frac{pi}{16}$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 3 '18 at 14:58









              MoKo19

              1914




              1914























                  3














                  Imagine $p$ and $q$ plotted along horizontal and vertical axes respectively. The given interval for $p$ and $q$ corresponds to a square of area 16. The discriminant is negative when $p^2 + q^2 lt 1$ which corresponds to a circle of area $pi$. The probability of non-real roots is the ratio of the area of the circle to the area of the square.






                  share|cite|improve this answer




























                    3














                    Imagine $p$ and $q$ plotted along horizontal and vertical axes respectively. The given interval for $p$ and $q$ corresponds to a square of area 16. The discriminant is negative when $p^2 + q^2 lt 1$ which corresponds to a circle of area $pi$. The probability of non-real roots is the ratio of the area of the circle to the area of the square.






                    share|cite|improve this answer


























                      3












                      3








                      3






                      Imagine $p$ and $q$ plotted along horizontal and vertical axes respectively. The given interval for $p$ and $q$ corresponds to a square of area 16. The discriminant is negative when $p^2 + q^2 lt 1$ which corresponds to a circle of area $pi$. The probability of non-real roots is the ratio of the area of the circle to the area of the square.






                      share|cite|improve this answer














                      Imagine $p$ and $q$ plotted along horizontal and vertical axes respectively. The given interval for $p$ and $q$ corresponds to a square of area 16. The discriminant is negative when $p^2 + q^2 lt 1$ which corresponds to a circle of area $pi$. The probability of non-real roots is the ratio of the area of the circle to the area of the square.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 3 '18 at 16:50

























                      answered Dec 3 '18 at 14:53









                      NickD

                      1,0051412




                      1,0051412























                          2














                          To nitpick:



                          I think the problem is incorrectly worded and we should be finding the probability of non-real roots: imaginary ones would be complex numbers $a+bi$ with $a=0.$



                          For the given equation, this only happens when $p=0$, so unless $p$ is chosen from finitely many values rather than from all the real numbers in $[-2,2]$, the probability is zero if we take imaginary literally.






                          share|cite|improve this answer




























                            2














                            To nitpick:



                            I think the problem is incorrectly worded and we should be finding the probability of non-real roots: imaginary ones would be complex numbers $a+bi$ with $a=0.$



                            For the given equation, this only happens when $p=0$, so unless $p$ is chosen from finitely many values rather than from all the real numbers in $[-2,2]$, the probability is zero if we take imaginary literally.






                            share|cite|improve this answer


























                              2












                              2








                              2






                              To nitpick:



                              I think the problem is incorrectly worded and we should be finding the probability of non-real roots: imaginary ones would be complex numbers $a+bi$ with $a=0.$



                              For the given equation, this only happens when $p=0$, so unless $p$ is chosen from finitely many values rather than from all the real numbers in $[-2,2]$, the probability is zero if we take imaginary literally.






                              share|cite|improve this answer














                              To nitpick:



                              I think the problem is incorrectly worded and we should be finding the probability of non-real roots: imaginary ones would be complex numbers $a+bi$ with $a=0.$



                              For the given equation, this only happens when $p=0$, so unless $p$ is chosen from finitely many values rather than from all the real numbers in $[-2,2]$, the probability is zero if we take imaginary literally.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Dec 3 '18 at 15:42

























                              answered Dec 3 '18 at 15:16









                              timtfj

                              1,026318




                              1,026318























                                  0














                                  Yes, you have to use the discriminant $D(p,q)$ of that polynomial. And you have to prove that if $A$ is the area of the region$$left{(p,q)in[-2,2]times[-2,2],middle|,D(p,q)<0right},$$then$$frac A{16}=fracpi4,$$where that $16$ is the area of the square $[-2,2]times[-2,2]$.






                                  share|cite|improve this answer


























                                    0














                                    Yes, you have to use the discriminant $D(p,q)$ of that polynomial. And you have to prove that if $A$ is the area of the region$$left{(p,q)in[-2,2]times[-2,2],middle|,D(p,q)<0right},$$then$$frac A{16}=fracpi4,$$where that $16$ is the area of the square $[-2,2]times[-2,2]$.






                                    share|cite|improve this answer
























                                      0












                                      0








                                      0






                                      Yes, you have to use the discriminant $D(p,q)$ of that polynomial. And you have to prove that if $A$ is the area of the region$$left{(p,q)in[-2,2]times[-2,2],middle|,D(p,q)<0right},$$then$$frac A{16}=fracpi4,$$where that $16$ is the area of the square $[-2,2]times[-2,2]$.






                                      share|cite|improve this answer












                                      Yes, you have to use the discriminant $D(p,q)$ of that polynomial. And you have to prove that if $A$ is the area of the region$$left{(p,q)in[-2,2]times[-2,2],middle|,D(p,q)<0right},$$then$$frac A{16}=fracpi4,$$where that $16$ is the area of the square $[-2,2]times[-2,2]$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 3 '18 at 14:51









                                      José Carlos Santos

                                      150k22122223




                                      150k22122223






























                                          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.





                                          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                          Please pay close attention to the following guidance:


                                          • 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.


                                          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%2f3024149%2ffind-the-probability-of-quadratic-having-imaginary-roots%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

                                          Berounka

                                          Image
                                          Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
                                          Read more

                                          Sphinx de Gizeh

                                          Image
                                          Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
                                          Read more

                                          Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

                                          Image
                                          3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
                                          Read more