Find The Probability of Quadratic having imaginary roots
$begingroup$
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
asked Dec 3 '18 at 14:45
SolvingTraineeSolvingTrainee
264
$endgroup$
add a comment |
$begingroup$
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
asked Dec 3 '18 at 14:45
SolvingTraineeSolvingTrainee
264
$endgroup$
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
add a comment |
$begingroup$
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
asked Dec 3 '18 at 14:45
SolvingTraineeSolvingTrainee
264
$endgroup$
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
asked Dec 3 '18 at 14:45
SolvingTraineeSolvingTrainee
264
asked Dec 3 '18 at 14:45
SolvingTraineeSolvingTrainee
264
asked Dec 3 '18 at 14:45
SolvingTraineeSolvingTrainee
264
asked Dec 3 '18 at 14:45
SolvingTraineeSolvingTrainee
264
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
add a comment |
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
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
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 $frac{pi}{16}$
answered Dec 3 '18 at 14:58
MoKo19MoKo19
1914
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 3 '18 at 14:53
NickDNickD
1,1211512
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 3 '18 at 15:16
timtfjtimtfj
2,448420
$endgroup$
add a comment |
$begingroup$
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 $[-2,2]times[-2,2]$.
answered Dec 3 '18 at 14:51
José Carlos SantosJosé Carlos Santos
163k22130234
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
$begingroup$
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 $frac{pi}{16}$
answered Dec 3 '18 at 14:58
MoKo19MoKo19
1914
$endgroup$
add a comment |
$begingroup$
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 $frac{pi}{16}$
answered Dec 3 '18 at 14:58
MoKo19MoKo19
1914
$endgroup$
add a comment |
$begingroup$
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 $frac{pi}{16}$
answered Dec 3 '18 at 14:58
MoKo19MoKo19
1914
$endgroup$
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 $frac{pi}{16}$
answered Dec 3 '18 at 14:58
MoKo19MoKo19
1914
answered Dec 3 '18 at 14:58
MoKo19MoKo19
1914
answered Dec 3 '18 at 14:58
MoKo19MoKo19
1914
answered Dec 3 '18 at 14:58
MoKo19MoKo19
1914
1914
add a comment |
add a comment |
$begingroup$
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.
answered Dec 3 '18 at 14:53
NickDNickD
1,1211512
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 3 '18 at 14:53
NickDNickD
1,1211512
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 3 '18 at 14:53
NickDNickD
1,1211512
$endgroup$
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.
answered Dec 3 '18 at 14:53
NickDNickD
1,1211512
edited Dec 3 '18 at 16:50
answered Dec 3 '18 at 14:53
NickDNickD
1,1211512
answered Dec 3 '18 at 14:53
NickDNickD
1,1211512
answered Dec 3 '18 at 14:53
NickDNickD
1,1211512
1,1211512
add a comment |
add a comment |
$begingroup$
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.
answered Dec 3 '18 at 15:16
timtfjtimtfj
2,448420
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 3 '18 at 15:16
timtfjtimtfj
2,448420
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 3 '18 at 15:16
timtfjtimtfj
2,448420
$endgroup$
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.
answered Dec 3 '18 at 15:16
timtfjtimtfj
2,448420
edited Dec 3 '18 at 15:42
answered Dec 3 '18 at 15:16
timtfjtimtfj
2,448420
answered Dec 3 '18 at 15:16
timtfjtimtfj
2,448420
answered Dec 3 '18 at 15:16
timtfjtimtfj
2,448420
2,448420
add a comment |
add a comment |
$begingroup$
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 $[-2,2]times[-2,2]$.
answered Dec 3 '18 at 14:51
José Carlos SantosJosé Carlos Santos
163k22130234
$endgroup$
add a comment |
$begingroup$
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 $[-2,2]times[-2,2]$.
answered Dec 3 '18 at 14:51
José Carlos SantosJosé Carlos Santos
163k22130234
$endgroup$
add a comment |
$begingroup$
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 $[-2,2]times[-2,2]$.
answered Dec 3 '18 at 14:51
José Carlos SantosJosé Carlos Santos
163k22130234
$endgroup$
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 $[-2,2]times[-2,2]$.
answered Dec 3 '18 at 14:51
José Carlos SantosJosé Carlos Santos
163k22130234
answered Dec 3 '18 at 14:51
José Carlos SantosJosé Carlos Santos
163k22130234
answered Dec 3 '18 at 14:51
José Carlos SantosJosé Carlos Santos
163k22130234
answered Dec 3 '18 at 14:51
José Carlos SantosJosé Carlos Santos
163k22130234
163k22130234
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
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