Mixing
Why does Wolfram Alpha say the roots of a cubic involve square roots of negative numbers, when all three...
$begingroup$
Can someone please explain Wolfram Alpha's response to the equation $x^3-3x-1=0$? The roots are real and graphic Wolfram itself shows them (they are approximately equal to $-1.5321, -0.3473$ and $1.8794$). However the roots explicitly given by Wolfram seem not to be real.
roots math-software
edited Nov 1 '16 at 9:06
Silverfish
9951123
asked Oct 31 '16 at 20:39
PiquitoPiquito
17.9k31437
$endgroup$
|
show 3 more comments
$begingroup$
Can someone please explain Wolfram Alpha's response to the equation $x^3-3x-1=0$? The roots are real and graphic Wolfram itself shows them (they are approximately equal to $-1.5321, -0.3473$ and $1.8794$). However the roots explicitly given by Wolfram seem not to be real.
roots math-software
edited Nov 1 '16 at 9:06
Silverfish
9951123
asked Oct 31 '16 at 20:39
PiquitoPiquito
17.9k31437
$endgroup$
28
$begingroup$
The imaginary bits cancel out.
$endgroup$
– Doug M
Oct 31 '16 at 20:41
4
$begingroup$
Please write a more descriptive title, thanks!
$endgroup$
– Nicolas Raoul
Nov 1 '16 at 7:06
6
$begingroup$
@DougM Any insight as to why Wolfram Alpha wouldn't simplify the equations if the imaginary bits do cancel out?
$endgroup$
– Ian Boyd
Nov 1 '16 at 12:47
7
$begingroup$
@IanBoyd That's the big plot twist in the story of the cubic formula: the question only involves real numbers, and the answer is real, but to write the answer using radicals, you have to use some complex intermediates. The fact that the imaginary parts add up to 0 and the end doesn't mean there's some way of rearranging the expression to get rid of them. If you try, you mess up the real part, which you need.
$endgroup$
– Wumpus Q. Wumbley
Nov 2 '16 at 1:03
1
$begingroup$
@WumpusQ.Wumbley #unlesstrig
$endgroup$
– Simply Beautiful Art
Nov 2 '16 at 11:35
|
show 3 more comments
$begingroup$
Can someone please explain Wolfram Alpha's response to the equation $x^3-3x-1=0$? The roots are real and graphic Wolfram itself shows them (they are approximately equal to $-1.5321, -0.3473$ and $1.8794$). However the roots explicitly given by Wolfram seem not to be real.
roots math-software
edited Nov 1 '16 at 9:06
Silverfish
9951123
asked Oct 31 '16 at 20:39
PiquitoPiquito
17.9k31437
$endgroup$
Can someone please explain Wolfram Alpha's response to the equation $x^3-3x-1=0$? The roots are real and graphic Wolfram itself shows them (they are approximately equal to $-1.5321, -0.3473$ and $1.8794$). However the roots explicitly given by Wolfram seem not to be real.
roots math-software
roots math-software
edited Nov 1 '16 at 9:06
Silverfish
9951123
asked Oct 31 '16 at 20:39
PiquitoPiquito
17.9k31437
edited Nov 1 '16 at 9:06
Silverfish
9951123
asked Oct 31 '16 at 20:39
PiquitoPiquito
17.9k31437
edited Nov 1 '16 at 9:06
Silverfish
9951123
edited Nov 1 '16 at 9:06
Silverfish
9951123
edited Nov 1 '16 at 9:06
Silverfish
9951123
9951123
asked Oct 31 '16 at 20:39
PiquitoPiquito
17.9k31437
asked Oct 31 '16 at 20:39
PiquitoPiquito
17.9k31437
asked Oct 31 '16 at 20:39
PiquitoPiquito
17.9k31437
17.9k31437
28
$begingroup$
The imaginary bits cancel out.
$endgroup$
– Doug M
Oct 31 '16 at 20:41
4
$begingroup$
Please write a more descriptive title, thanks!
$endgroup$
– Nicolas Raoul
Nov 1 '16 at 7:06
6
$begingroup$
@DougM Any insight as to why Wolfram Alpha wouldn't simplify the equations if the imaginary bits do cancel out?
$endgroup$
– Ian Boyd
Nov 1 '16 at 12:47
7
$begingroup$
@IanBoyd That's the big plot twist in the story of the cubic formula: the question only involves real numbers, and the answer is real, but to write the answer using radicals, you have to use some complex intermediates. The fact that the imaginary parts add up to 0 and the end doesn't mean there's some way of rearranging the expression to get rid of them. If you try, you mess up the real part, which you need.
$endgroup$
– Wumpus Q. Wumbley
Nov 2 '16 at 1:03
1
$begingroup$
@WumpusQ.Wumbley #unlesstrig
$endgroup$
– Simply Beautiful Art
Nov 2 '16 at 11:35
|
show 3 more comments
28
$begingroup$
The imaginary bits cancel out.
$endgroup$
– Doug M
Oct 31 '16 at 20:41
4
$begingroup$
Please write a more descriptive title, thanks!
$endgroup$
– Nicolas Raoul
Nov 1 '16 at 7:06
6
$begingroup$
@DougM Any insight as to why Wolfram Alpha wouldn't simplify the equations if the imaginary bits do cancel out?
$endgroup$
– Ian Boyd
Nov 1 '16 at 12:47
7
$begingroup$
@IanBoyd That's the big plot twist in the story of the cubic formula: the question only involves real numbers, and the answer is real, but to write the answer using radicals, you have to use some complex intermediates. The fact that the imaginary parts add up to 0 and the end doesn't mean there's some way of rearranging the expression to get rid of them. If you try, you mess up the real part, which you need.
$endgroup$
– Wumpus Q. Wumbley
Nov 2 '16 at 1:03
1
$begingroup$
@WumpusQ.Wumbley #unlesstrig
$endgroup$
– Simply Beautiful Art
Nov 2 '16 at 11:35
28
28
$begingroup$
The imaginary bits cancel out.
$endgroup$
– Doug M
Oct 31 '16 at 20:41
$begingroup$
The imaginary bits cancel out.
$endgroup$
– Doug M
Oct 31 '16 at 20:41
4
4
$begingroup$
Please write a more descriptive title, thanks!
$endgroup$
– Nicolas Raoul
Nov 1 '16 at 7:06
$begingroup$
Please write a more descriptive title, thanks!
$endgroup$
– Nicolas Raoul
Nov 1 '16 at 7:06
6
6
$begingroup$
@DougM Any insight as to why Wolfram Alpha wouldn't simplify the equations if the imaginary bits do cancel out?
$endgroup$
– Ian Boyd
Nov 1 '16 at 12:47
$begingroup$
@DougM Any insight as to why Wolfram Alpha wouldn't simplify the equations if the imaginary bits do cancel out?
$endgroup$
– Ian Boyd
Nov 1 '16 at 12:47
7
7
$begingroup$
@IanBoyd That's the big plot twist in the story of the cubic formula: the question only involves real numbers, and the answer is real, but to write the answer using radicals, you have to use some complex intermediates. The fact that the imaginary parts add up to 0 and the end doesn't mean there's some way of rearranging the expression to get rid of them. If you try, you mess up the real part, which you need.
$endgroup$
– Wumpus Q. Wumbley
Nov 2 '16 at 1:03
$begingroup$
@IanBoyd That's the big plot twist in the story of the cubic formula: the question only involves real numbers, and the answer is real, but to write the answer using radicals, you have to use some complex intermediates. The fact that the imaginary parts add up to 0 and the end doesn't mean there's some way of rearranging the expression to get rid of them. If you try, you mess up the real part, which you need.
$endgroup$
– Wumpus Q. Wumbley
Nov 2 '16 at 1:03
1
1
$begingroup$
@WumpusQ.Wumbley #unlesstrig
$endgroup$
– Simply Beautiful Art
Nov 2 '16 at 11:35
$begingroup$
@WumpusQ.Wumbley #unlesstrig
$endgroup$
– Simply Beautiful Art
Nov 2 '16 at 11:35
|
show 3 more comments
4 Answers
4
active
oldest
votes
$begingroup$
When the discriminant of the equation is negative, there are $3$ real roots. It's precisely in this case that Cardano's formulæ do not work, because the radicands are negative.
It is even historically the reason why square roots of negative numbers were introduced: the formulae become valid in all cases.
answered Oct 31 '16 at 20:45
BernardBernard
119k740113
$endgroup$
13
$begingroup$
I feel like the wording "Cardano's formulae do not work" is somewhat misleading - they do work, but you have to think about complex numbers at this point.
$endgroup$
– Wojowu
Nov 1 '16 at 12:30
1
$begingroup$
That's what I meant: their plain use doesn't work. I think this is clear from the following comment.
$endgroup$
– Bernard
Nov 1 '16 at 13:56
add a comment |
$begingroup$
This is an example of the casus irreducibilis: the formulas for solving cubics need complex numbers when the cubic has three real roots.
answered Oct 31 '16 at 20:42
lhflhf
163k10169393
$endgroup$
1
$begingroup$
I knew that my God! Why I did not remember it?
$endgroup$
– Piquito
Oct 31 '16 at 21:03
add a comment |
$begingroup$
For your interest, the real representation of the roots is given by
$$x_k=2cosleft(fracpi9+frac{2pi k}3right)qquad k=0,1,2$$
I always find these trigonometric representations nicer.
This comes directly from the following identity:
$$cos(3a)=4cos^3(a)-3cos(a)$$
$$cos(3arccos(a))=4a^3-3a$$
And lastly the fact that cosine is periodic.
answered Oct 31 '16 at 23:48
Simply Beautiful ArtSimply Beautiful Art
50.5k578181
$endgroup$
$begingroup$
Absolutely. And it is known too. Regards.
$endgroup$
– Piquito
Nov 1 '16 at 10:48
$begingroup$
Could you add a few words on how you calculated these roots?
$endgroup$
– Strants
Nov 1 '16 at 15:35
$begingroup$
@Strants there you go. The trick to this particular problem was the substitution $x=2y$, then dividing both sides by $2$. Then everything is in the perfect form.
$endgroup$
– Simply Beautiful Art
Nov 1 '16 at 16:42
$begingroup$
How in the world did you know that identity I wish to be able to recognize this like you.
$endgroup$
– Ahmed S. Attaalla
Jan 8 '17 at 4:27
$begingroup$
@AhmedS.Attaalla It's just my favorite way to factor cubic polynomials :D Comes with practice mate.
$endgroup$
– Simply Beautiful Art
Jan 8 '17 at 12:54
add a comment |
$begingroup$
Solving equation $x^3-3x-1=0$ with free CAS Maxima:
answered Sep 11 '18 at 7:47
Aleksas DomarkasAleksas Domarkas
8996
$endgroup$
$begingroup$
What issolvet? Solve using trig?
$endgroup$
– GEdgar
Sep 11 '18 at 12:48
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%2f1993570%2fwhy-does-wolfram-alpha-say-the-roots-of-a-cubic-involve-square-roots-of-negative%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
$begingroup$
When the discriminant of the equation is negative, there are $3$ real roots. It's precisely in this case that Cardano's formulæ do not work, because the radicands are negative.
It is even historically the reason why square roots of negative numbers were introduced: the formulae become valid in all cases.
answered Oct 31 '16 at 20:45
BernardBernard
119k740113
$endgroup$
13
$begingroup$
I feel like the wording "Cardano's formulae do not work" is somewhat misleading - they do work, but you have to think about complex numbers at this point.
$endgroup$
– Wojowu
Nov 1 '16 at 12:30
1
$begingroup$
That's what I meant: their plain use doesn't work. I think this is clear from the following comment.
$endgroup$
– Bernard
Nov 1 '16 at 13:56
add a comment |
$begingroup$
When the discriminant of the equation is negative, there are $3$ real roots. It's precisely in this case that Cardano's formulæ do not work, because the radicands are negative.
It is even historically the reason why square roots of negative numbers were introduced: the formulae become valid in all cases.
answered Oct 31 '16 at 20:45
BernardBernard
119k740113
$endgroup$
13
$begingroup$
I feel like the wording "Cardano's formulae do not work" is somewhat misleading - they do work, but you have to think about complex numbers at this point.
$endgroup$
– Wojowu
Nov 1 '16 at 12:30
1
$begingroup$
That's what I meant: their plain use doesn't work. I think this is clear from the following comment.
$endgroup$
– Bernard
Nov 1 '16 at 13:56
add a comment |
$begingroup$
When the discriminant of the equation is negative, there are $3$ real roots. It's precisely in this case that Cardano's formulæ do not work, because the radicands are negative.
It is even historically the reason why square roots of negative numbers were introduced: the formulae become valid in all cases.
answered Oct 31 '16 at 20:45
BernardBernard
119k740113
$endgroup$
When the discriminant of the equation is negative, there are $3$ real roots. It's precisely in this case that Cardano's formulæ do not work, because the radicands are negative.
It is even historically the reason why square roots of negative numbers were introduced: the formulae become valid in all cases.
answered Oct 31 '16 at 20:45
BernardBernard
119k740113
edited Jan 4 at 22:32
answered Oct 31 '16 at 20:45
BernardBernard
119k740113
answered Oct 31 '16 at 20:45
BernardBernard
119k740113
answered Oct 31 '16 at 20:45
BernardBernard
119k740113
119k740113
13
$begingroup$
I feel like the wording "Cardano's formulae do not work" is somewhat misleading - they do work, but you have to think about complex numbers at this point.
$endgroup$
– Wojowu
Nov 1 '16 at 12:30
1
$begingroup$
That's what I meant: their plain use doesn't work. I think this is clear from the following comment.
$endgroup$
– Bernard
Nov 1 '16 at 13:56
add a comment |
13
$begingroup$
I feel like the wording "Cardano's formulae do not work" is somewhat misleading - they do work, but you have to think about complex numbers at this point.
$endgroup$
– Wojowu
Nov 1 '16 at 12:30
1
$begingroup$
That's what I meant: their plain use doesn't work. I think this is clear from the following comment.
$endgroup$
– Bernard
Nov 1 '16 at 13:56
13
13
$begingroup$
I feel like the wording "Cardano's formulae do not work" is somewhat misleading - they do work, but you have to think about complex numbers at this point.
$endgroup$
– Wojowu
Nov 1 '16 at 12:30
$begingroup$
I feel like the wording "Cardano's formulae do not work" is somewhat misleading - they do work, but you have to think about complex numbers at this point.
$endgroup$
– Wojowu
Nov 1 '16 at 12:30
1
1
$begingroup$
That's what I meant: their plain use doesn't work. I think this is clear from the following comment.
$endgroup$
– Bernard
Nov 1 '16 at 13:56
$begingroup$
That's what I meant: their plain use doesn't work. I think this is clear from the following comment.
$endgroup$
– Bernard
Nov 1 '16 at 13:56
add a comment |
$begingroup$
This is an example of the casus irreducibilis: the formulas for solving cubics need complex numbers when the cubic has three real roots.
answered Oct 31 '16 at 20:42
lhflhf
163k10169393
$endgroup$
1
$begingroup$
I knew that my God! Why I did not remember it?
$endgroup$
– Piquito
Oct 31 '16 at 21:03
add a comment |
$begingroup$
This is an example of the casus irreducibilis: the formulas for solving cubics need complex numbers when the cubic has three real roots.
answered Oct 31 '16 at 20:42
lhflhf
163k10169393
$endgroup$
1
$begingroup$
I knew that my God! Why I did not remember it?
$endgroup$
– Piquito
Oct 31 '16 at 21:03
add a comment |
$begingroup$
This is an example of the casus irreducibilis: the formulas for solving cubics need complex numbers when the cubic has three real roots.
answered Oct 31 '16 at 20:42
lhflhf
163k10169393
$endgroup$
This is an example of the casus irreducibilis: the formulas for solving cubics need complex numbers when the cubic has three real roots.
answered Oct 31 '16 at 20:42
lhflhf
163k10169393
answered Oct 31 '16 at 20:42
lhflhf
163k10169393
answered Oct 31 '16 at 20:42
lhflhf
163k10169393
answered Oct 31 '16 at 20:42
lhflhf
163k10169393
163k10169393
1
$begingroup$
I knew that my God! Why I did not remember it?
$endgroup$
– Piquito
Oct 31 '16 at 21:03
add a comment |
1
$begingroup$
I knew that my God! Why I did not remember it?
$endgroup$
– Piquito
Oct 31 '16 at 21:03
1
1
$begingroup$
I knew that my God! Why I did not remember it?
$endgroup$
– Piquito
Oct 31 '16 at 21:03
$begingroup$
I knew that my God! Why I did not remember it?
$endgroup$
– Piquito
Oct 31 '16 at 21:03
add a comment |
$begingroup$
For your interest, the real representation of the roots is given by
$$x_k=2cosleft(fracpi9+frac{2pi k}3right)qquad k=0,1,2$$
I always find these trigonometric representations nicer.
This comes directly from the following identity:
$$cos(3a)=4cos^3(a)-3cos(a)$$
$$cos(3arccos(a))=4a^3-3a$$
And lastly the fact that cosine is periodic.
answered Oct 31 '16 at 23:48
Simply Beautiful ArtSimply Beautiful Art
50.5k578181
$endgroup$
$begingroup$
Absolutely. And it is known too. Regards.
$endgroup$
– Piquito
Nov 1 '16 at 10:48
$begingroup$
Could you add a few words on how you calculated these roots?
$endgroup$
– Strants
Nov 1 '16 at 15:35
$begingroup$
@Strants there you go. The trick to this particular problem was the substitution $x=2y$, then dividing both sides by $2$. Then everything is in the perfect form.
$endgroup$
– Simply Beautiful Art
Nov 1 '16 at 16:42
$begingroup$
How in the world did you know that identity I wish to be able to recognize this like you.
$endgroup$
– Ahmed S. Attaalla
Jan 8 '17 at 4:27
$begingroup$
@AhmedS.Attaalla It's just my favorite way to factor cubic polynomials :D Comes with practice mate.
$endgroup$
– Simply Beautiful Art
Jan 8 '17 at 12:54
add a comment |
$begingroup$
For your interest, the real representation of the roots is given by
$$x_k=2cosleft(fracpi9+frac{2pi k}3right)qquad k=0,1,2$$
I always find these trigonometric representations nicer.
This comes directly from the following identity:
$$cos(3a)=4cos^3(a)-3cos(a)$$
$$cos(3arccos(a))=4a^3-3a$$
And lastly the fact that cosine is periodic.
answered Oct 31 '16 at 23:48
Simply Beautiful ArtSimply Beautiful Art
50.5k578181
$endgroup$
$begingroup$
Absolutely. And it is known too. Regards.
$endgroup$
– Piquito
Nov 1 '16 at 10:48
$begingroup$
Could you add a few words on how you calculated these roots?
$endgroup$
– Strants
Nov 1 '16 at 15:35
$begingroup$
@Strants there you go. The trick to this particular problem was the substitution $x=2y$, then dividing both sides by $2$. Then everything is in the perfect form.
$endgroup$
– Simply Beautiful Art
Nov 1 '16 at 16:42
$begingroup$
How in the world did you know that identity I wish to be able to recognize this like you.
$endgroup$
– Ahmed S. Attaalla
Jan 8 '17 at 4:27
$begingroup$
@AhmedS.Attaalla It's just my favorite way to factor cubic polynomials :D Comes with practice mate.
$endgroup$
– Simply Beautiful Art
Jan 8 '17 at 12:54
add a comment |
$begingroup$
For your interest, the real representation of the roots is given by
$$x_k=2cosleft(fracpi9+frac{2pi k}3right)qquad k=0,1,2$$
I always find these trigonometric representations nicer.
This comes directly from the following identity:
$$cos(3a)=4cos^3(a)-3cos(a)$$
$$cos(3arccos(a))=4a^3-3a$$
And lastly the fact that cosine is periodic.
answered Oct 31 '16 at 23:48
Simply Beautiful ArtSimply Beautiful Art
50.5k578181
$endgroup$
For your interest, the real representation of the roots is given by
$$x_k=2cosleft(fracpi9+frac{2pi k}3right)qquad k=0,1,2$$
I always find these trigonometric representations nicer.
This comes directly from the following identity:
$$cos(3a)=4cos^3(a)-3cos(a)$$
$$cos(3arccos(a))=4a^3-3a$$
And lastly the fact that cosine is periodic.
answered Oct 31 '16 at 23:48
Simply Beautiful ArtSimply Beautiful Art
50.5k578181
edited Nov 1 '16 at 16:40
answered Oct 31 '16 at 23:48
Simply Beautiful ArtSimply Beautiful Art
50.5k578181
answered Oct 31 '16 at 23:48
Simply Beautiful ArtSimply Beautiful Art
50.5k578181
answered Oct 31 '16 at 23:48
Simply Beautiful ArtSimply Beautiful Art
50.5k578181
50.5k578181
$begingroup$
Absolutely. And it is known too. Regards.
$endgroup$
– Piquito
Nov 1 '16 at 10:48
$begingroup$
Could you add a few words on how you calculated these roots?
$endgroup$
– Strants
Nov 1 '16 at 15:35
$begingroup$
@Strants there you go. The trick to this particular problem was the substitution $x=2y$, then dividing both sides by $2$. Then everything is in the perfect form.
$endgroup$
– Simply Beautiful Art
Nov 1 '16 at 16:42
$begingroup$
How in the world did you know that identity I wish to be able to recognize this like you.
$endgroup$
– Ahmed S. Attaalla
Jan 8 '17 at 4:27
$begingroup$
@AhmedS.Attaalla It's just my favorite way to factor cubic polynomials :D Comes with practice mate.
$endgroup$
– Simply Beautiful Art
Jan 8 '17 at 12:54
add a comment |
$begingroup$
Absolutely. And it is known too. Regards.
$endgroup$
– Piquito
Nov 1 '16 at 10:48
$begingroup$
Could you add a few words on how you calculated these roots?
$endgroup$
– Strants
Nov 1 '16 at 15:35
$begingroup$
@Strants there you go. The trick to this particular problem was the substitution $x=2y$, then dividing both sides by $2$. Then everything is in the perfect form.
$endgroup$
– Simply Beautiful Art
Nov 1 '16 at 16:42
$begingroup$
How in the world did you know that identity I wish to be able to recognize this like you.
$endgroup$
– Ahmed S. Attaalla
Jan 8 '17 at 4:27
$begingroup$
@AhmedS.Attaalla It's just my favorite way to factor cubic polynomials :D Comes with practice mate.
$endgroup$
– Simply Beautiful Art
Jan 8 '17 at 12:54
$begingroup$
Absolutely. And it is known too. Regards.
$endgroup$
– Piquito
Nov 1 '16 at 10:48
$begingroup$
Absolutely. And it is known too. Regards.
$endgroup$
– Piquito
Nov 1 '16 at 10:48
$begingroup$
Could you add a few words on how you calculated these roots?
$endgroup$
– Strants
Nov 1 '16 at 15:35
$begingroup$
Could you add a few words on how you calculated these roots?
$endgroup$
– Strants
Nov 1 '16 at 15:35
$begingroup$
@Strants there you go. The trick to this particular problem was the substitution $x=2y$, then dividing both sides by $2$. Then everything is in the perfect form.
$endgroup$
– Simply Beautiful Art
Nov 1 '16 at 16:42
$begingroup$
@Strants there you go. The trick to this particular problem was the substitution $x=2y$, then dividing both sides by $2$. Then everything is in the perfect form.
$endgroup$
– Simply Beautiful Art
Nov 1 '16 at 16:42
$begingroup$
How in the world did you know that identity I wish to be able to recognize this like you.
$endgroup$
– Ahmed S. Attaalla
Jan 8 '17 at 4:27
$begingroup$
How in the world did you know that identity I wish to be able to recognize this like you.
$endgroup$
– Ahmed S. Attaalla
Jan 8 '17 at 4:27
$begingroup$
@AhmedS.Attaalla It's just my favorite way to factor cubic polynomials :D Comes with practice mate.
$endgroup$
– Simply Beautiful Art
Jan 8 '17 at 12:54
$begingroup$
@AhmedS.Attaalla It's just my favorite way to factor cubic polynomials :D Comes with practice mate.
$endgroup$
– Simply Beautiful Art
Jan 8 '17 at 12:54
add a comment |
$begingroup$
Solving equation $x^3-3x-1=0$ with free CAS Maxima:
answered Sep 11 '18 at 7:47
Aleksas DomarkasAleksas Domarkas
8996
$endgroup$
$begingroup$
What issolvet? Solve using trig?
$endgroup$
– GEdgar
Sep 11 '18 at 12:48
add a comment |
$begingroup$
Solving equation $x^3-3x-1=0$ with free CAS Maxima:
answered Sep 11 '18 at 7:47
Aleksas DomarkasAleksas Domarkas
8996
$endgroup$
$begingroup$
What issolvet? Solve using trig?
$endgroup$
– GEdgar
Sep 11 '18 at 12:48
add a comment |
$begingroup$
Solving equation $x^3-3x-1=0$ with free CAS Maxima:
answered Sep 11 '18 at 7:47
Aleksas DomarkasAleksas Domarkas
8996
$endgroup$
Solving equation $x^3-3x-1=0$ with free CAS Maxima:
answered Sep 11 '18 at 7:47
Aleksas DomarkasAleksas Domarkas
8996
answered Sep 11 '18 at 7:47
Aleksas DomarkasAleksas Domarkas
8996
answered Sep 11 '18 at 7:47
Aleksas DomarkasAleksas Domarkas
8996
answered Sep 11 '18 at 7:47
Aleksas DomarkasAleksas Domarkas
8996
8996
$begingroup$
What issolvet? Solve using trig?
$endgroup$
– GEdgar
Sep 11 '18 at 12:48
add a comment |
$begingroup$
What issolvet? Solve using trig?
$endgroup$
– GEdgar
Sep 11 '18 at 12:48
$begingroup$
What is
solvet? Solve using trig?$endgroup$
– GEdgar
Sep 11 '18 at 12:48
$begingroup$
What is
solvet? Solve using trig?$endgroup$
– GEdgar
Sep 11 '18 at 12:48
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%2f1993570%2fwhy-does-wolfram-alpha-say-the-roots-of-a-cubic-involve-square-roots-of-negative%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
28
$begingroup$
The imaginary bits cancel out.
$endgroup$
– Doug M
Oct 31 '16 at 20:41
4
$begingroup$
Please write a more descriptive title, thanks!
$endgroup$
– Nicolas Raoul
Nov 1 '16 at 7:06
6
$begingroup$
@DougM Any insight as to why Wolfram Alpha wouldn't simplify the equations if the imaginary bits do cancel out?
$endgroup$
– Ian Boyd
Nov 1 '16 at 12:47
7
$begingroup$
@IanBoyd That's the big plot twist in the story of the cubic formula: the question only involves real numbers, and the answer is real, but to write the answer using radicals, you have to use some complex intermediates. The fact that the imaginary parts add up to 0 and the end doesn't mean there's some way of rearranging the expression to get rid of them. If you try, you mess up the real part, which you need.
$endgroup$
– Wumpus Q. Wumbley
Nov 2 '16 at 1:03
1
$begingroup$
@WumpusQ.Wumbley #unlesstrig
$endgroup$
– Simply Beautiful Art
Nov 2 '16 at 11:35