Find The Probability of Quadratic having imaginary roots












1












$begingroup$


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







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$endgroup$








  • 1




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










  • $begingroup$
    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.
    $endgroup$
    – NickD
    Dec 3 '18 at 15:30


















1












$begingroup$


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









$endgroup$








  • 1




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










  • $begingroup$
    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.
    $endgroup$
    – NickD
    Dec 3 '18 at 15:30
















1












1








1





$begingroup$


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









$endgroup$




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






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asked Dec 3 '18 at 14:45









SolvingTraineeSolvingTrainee

264




264








  • 1




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










  • $begingroup$
    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.
    $endgroup$
    – NickD
    Dec 3 '18 at 15:30
















  • 1




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










  • $begingroup$
    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.
    $endgroup$
    – NickD
    Dec 3 '18 at 15:30










1




1




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




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












$begingroup$
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.
$endgroup$
– NickD
Dec 3 '18 at 15:30






$begingroup$
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.
$endgroup$
– NickD
Dec 3 '18 at 15:30












4 Answers
4






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3












$begingroup$

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









$endgroup$





















    3












    $begingroup$

    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











    $endgroup$





















      2












      $begingroup$

      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











      $endgroup$





















        0












        $begingroup$

        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









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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

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          active

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          3












          $begingroup$

          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









          $endgroup$


















            3












            $begingroup$

            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









            $endgroup$
















              3












              3








              3





              $begingroup$

              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









              $endgroup$



              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









              MoKo19MoKo19

              1914




              1914























                  3












                  $begingroup$

                  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











                  $endgroup$


















                    3












                    $begingroup$

                    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











                    $endgroup$
















                      3












                      3








                      3





                      $begingroup$

                      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











                      $endgroup$



                      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









                      NickDNickD

                      1,1211512




                      1,1211512























                          2












                          $begingroup$

                          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











                          $endgroup$


















                            2












                            $begingroup$

                            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











                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              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











                              $endgroup$



                              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









                              timtfjtimtfj

                              2,448420




                              2,448420























                                  0












                                  $begingroup$

                                  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









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    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









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      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









                                      $endgroup$



                                      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 SantosJosé Carlos Santos

                                      163k22130234




                                      163k22130234






























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