Do cubics always have one real root? [closed]
Clash Royale CLAN TAG#URR8PPP
$begingroup$
I've seen a few conflicting pieces of information online.
So far, I know that with real coefficients there will always be one real root. But how about with complex coefficients?
At very least could you give me a counterexample? A cubic with no real roots.
polynomials complex-numbers roots examples-counterexamples real-numbers
edited Mar 10 at 15:53
YuiTo Cheng
2,3084937
asked Mar 10 at 1:30
user7971589user7971589
273
$endgroup$
closed as off-topic by user21820, Alex Provost, Eevee Trainer, Carl Mummert, Parcly Taxel Mar 11 at 3:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Alex Provost, Eevee Trainer, Carl Mummert, Parcly Taxel
add a comment |
$begingroup$
I've seen a few conflicting pieces of information online.
So far, I know that with real coefficients there will always be one real root. But how about with complex coefficients?
At very least could you give me a counterexample? A cubic with no real roots.
polynomials complex-numbers roots examples-counterexamples real-numbers
edited Mar 10 at 15:53
YuiTo Cheng
2,3084937
asked Mar 10 at 1:30
user7971589user7971589
273
$endgroup$
closed as off-topic by user21820, Alex Provost, Eevee Trainer, Carl Mummert, Parcly Taxel Mar 11 at 3:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Alex Provost, Eevee Trainer, Carl Mummert, Parcly Taxel
5
$begingroup$
It's more clear to say "at least one real root" or "a real root". There are real cubics that have three real roots (there are none that have exactly two though).
$endgroup$
– quid♦
Mar 10 at 14:47
6
$begingroup$
@quid: If you count with multiplicity, that's true. If you don't, then you can have $f(x) = (x-1)(x-2)^2$.
$endgroup$
– Kevin
Mar 10 at 17:22
1
$begingroup$
Indeed @Kevin I should have made that explicity.
$endgroup$
– quid♦
Mar 10 at 19:15
add a comment |
$begingroup$
I've seen a few conflicting pieces of information online.
So far, I know that with real coefficients there will always be one real root. But how about with complex coefficients?
At very least could you give me a counterexample? A cubic with no real roots.
polynomials complex-numbers roots examples-counterexamples real-numbers
edited Mar 10 at 15:53
YuiTo Cheng
2,3084937
asked Mar 10 at 1:30
user7971589user7971589
273
$endgroup$
I've seen a few conflicting pieces of information online.
So far, I know that with real coefficients there will always be one real root. But how about with complex coefficients?
At very least could you give me a counterexample? A cubic with no real roots.
polynomials complex-numbers roots examples-counterexamples real-numbers
polynomials complex-numbers roots examples-counterexamples real-numbers
edited Mar 10 at 15:53
YuiTo Cheng
2,3084937
asked Mar 10 at 1:30
user7971589user7971589
273
edited Mar 10 at 15:53
YuiTo Cheng
2,3084937
asked Mar 10 at 1:30
user7971589user7971589
273
edited Mar 10 at 15:53
YuiTo Cheng
2,3084937
edited Mar 10 at 15:53
YuiTo Cheng
2,3084937
edited Mar 10 at 15:53
YuiTo Cheng
2,3084937
2,3084937
asked Mar 10 at 1:30
user7971589user7971589
273
asked Mar 10 at 1:30
user7971589user7971589
273
asked Mar 10 at 1:30
user7971589user7971589
273
273
closed as off-topic by user21820, Alex Provost, Eevee Trainer, Carl Mummert, Parcly Taxel Mar 11 at 3:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Alex Provost, Eevee Trainer, Carl Mummert, Parcly Taxel
closed as off-topic by user21820, Alex Provost, Eevee Trainer, Carl Mummert, Parcly Taxel Mar 11 at 3:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Alex Provost, Eevee Trainer, Carl Mummert, Parcly Taxel
5
$begingroup$
It's more clear to say "at least one real root" or "a real root". There are real cubics that have three real roots (there are none that have exactly two though).
$endgroup$
– quid♦
Mar 10 at 14:47
6
$begingroup$
@quid: If you count with multiplicity, that's true. If you don't, then you can have $f(x) = (x-1)(x-2)^2$.
$endgroup$
– Kevin
Mar 10 at 17:22
1
$begingroup$
Indeed @Kevin I should have made that explicity.
$endgroup$
– quid♦
Mar 10 at 19:15
add a comment |
5
$begingroup$
It's more clear to say "at least one real root" or "a real root". There are real cubics that have three real roots (there are none that have exactly two though).
$endgroup$
– quid♦
Mar 10 at 14:47
6
$begingroup$
@quid: If you count with multiplicity, that's true. If you don't, then you can have $f(x) = (x-1)(x-2)^2$.
$endgroup$
– Kevin
Mar 10 at 17:22
1
$begingroup$
Indeed @Kevin I should have made that explicity.
$endgroup$
– quid♦
Mar 10 at 19:15
5
5
$begingroup$
It's more clear to say "at least one real root" or "a real root". There are real cubics that have three real roots (there are none that have exactly two though).
$endgroup$
– quid♦
Mar 10 at 14:47
$begingroup$
It's more clear to say "at least one real root" or "a real root". There are real cubics that have three real roots (there are none that have exactly two though).
$endgroup$
– quid♦
Mar 10 at 14:47
6
6
$begingroup$
@quid: If you count with multiplicity, that's true. If you don't, then you can have $f(x) = (x-1)(x-2)^2$.
$endgroup$
– Kevin
Mar 10 at 17:22
$begingroup$
@quid: If you count with multiplicity, that's true. If you don't, then you can have $f(x) = (x-1)(x-2)^2$.
$endgroup$
– Kevin
Mar 10 at 17:22
1
1
$begingroup$
Indeed @Kevin I should have made that explicity.
$endgroup$
– quid♦
Mar 10 at 19:15
$begingroup$
Indeed @Kevin I should have made that explicity.
$endgroup$
– quid♦
Mar 10 at 19:15
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
One of the best things you can remember is that over a field (like the reals or complex numbers) roots come from linear factors. Use this to build your own examples: $f(z) =(z-i)^3$. If you want three distinct complex roots, do something like $f(z) = (z-i)(z+i)(z-2i)$.
answered Mar 10 at 1:31
RandallRandall
10.6k11431
$endgroup$
add a comment |
$begingroup$
As you already know, a cubic with real coefficients always has at least one real root, so there is no counterexample of a cubic with real coefficients with no real roots.
A cubic with complex coefficients with no real roots is easy to find; take $x^3+i$.
answered Mar 10 at 1:31
ServaesServaes
30k342101
$endgroup$
$begingroup$
@Glen_b It is reasonable to say has one X to mean there exists an X, though it would be more clear to say there is at least one X.
$endgroup$
– jgon
Mar 12 at 20:21
1
$begingroup$
@Glen_b I have edited to remove the ambiguity.
$endgroup$
– Servaes
Mar 12 at 20:22
add a comment |
$begingroup$
Over the complex numbers, every polynomial factors into roots. So we can take any cubic and write it as $a(x-u)(x-v)(x-w)$ where $u, v, w$ are the roots (they don't need to be distict) and $a$ is the leading coefficient. This lets us form polynomials with only complex roots such as $(x-i)^3$.
However, if all the original coefficients of the polynomial are real, and $c$ is a complex root, then its conjugate $barc$ must also be a root: complex roots to polynomials with real coeffecients must come in pairs. This is called the complex conjugate root theorem.
This means that a polynomial with real coefficients and odd degree will always have at least one real root, which answers the case for cubics. A quadratic with negative discriminant on the other hand has two, conjugate complex roots.
answered Mar 10 at 18:28
BristolBristol
1864
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One of the best things you can remember is that over a field (like the reals or complex numbers) roots come from linear factors. Use this to build your own examples: $f(z) =(z-i)^3$. If you want three distinct complex roots, do something like $f(z) = (z-i)(z+i)(z-2i)$.
answered Mar 10 at 1:31
RandallRandall
10.6k11431
$endgroup$
add a comment |
$begingroup$
One of the best things you can remember is that over a field (like the reals or complex numbers) roots come from linear factors. Use this to build your own examples: $f(z) =(z-i)^3$. If you want three distinct complex roots, do something like $f(z) = (z-i)(z+i)(z-2i)$.
answered Mar 10 at 1:31
RandallRandall
10.6k11431
$endgroup$
add a comment |
$begingroup$
One of the best things you can remember is that over a field (like the reals or complex numbers) roots come from linear factors. Use this to build your own examples: $f(z) =(z-i)^3$. If you want three distinct complex roots, do something like $f(z) = (z-i)(z+i)(z-2i)$.
answered Mar 10 at 1:31
RandallRandall
10.6k11431
$endgroup$
One of the best things you can remember is that over a field (like the reals or complex numbers) roots come from linear factors. Use this to build your own examples: $f(z) =(z-i)^3$. If you want three distinct complex roots, do something like $f(z) = (z-i)(z+i)(z-2i)$.
answered Mar 10 at 1:31
RandallRandall
10.6k11431
answered Mar 10 at 1:31
RandallRandall
10.6k11431
answered Mar 10 at 1:31
RandallRandall
10.6k11431
answered Mar 10 at 1:31
RandallRandall
10.6k11431
10.6k11431
add a comment |
add a comment |
$begingroup$
As you already know, a cubic with real coefficients always has at least one real root, so there is no counterexample of a cubic with real coefficients with no real roots.
A cubic with complex coefficients with no real roots is easy to find; take $x^3+i$.
answered Mar 10 at 1:31
ServaesServaes
30k342101
$endgroup$
$begingroup$
@Glen_b It is reasonable to say has one X to mean there exists an X, though it would be more clear to say there is at least one X.
$endgroup$
– jgon
Mar 12 at 20:21
1
$begingroup$
@Glen_b I have edited to remove the ambiguity.
$endgroup$
– Servaes
Mar 12 at 20:22
add a comment |
$begingroup$
As you already know, a cubic with real coefficients always has at least one real root, so there is no counterexample of a cubic with real coefficients with no real roots.
A cubic with complex coefficients with no real roots is easy to find; take $x^3+i$.
answered Mar 10 at 1:31
ServaesServaes
30k342101
$endgroup$
$begingroup$
@Glen_b It is reasonable to say has one X to mean there exists an X, though it would be more clear to say there is at least one X.
$endgroup$
– jgon
Mar 12 at 20:21
1
$begingroup$
@Glen_b I have edited to remove the ambiguity.
$endgroup$
– Servaes
Mar 12 at 20:22
add a comment |
$begingroup$
As you already know, a cubic with real coefficients always has at least one real root, so there is no counterexample of a cubic with real coefficients with no real roots.
A cubic with complex coefficients with no real roots is easy to find; take $x^3+i$.
answered Mar 10 at 1:31
ServaesServaes
30k342101
$endgroup$
As you already know, a cubic with real coefficients always has at least one real root, so there is no counterexample of a cubic with real coefficients with no real roots.
A cubic with complex coefficients with no real roots is easy to find; take $x^3+i$.
answered Mar 10 at 1:31
ServaesServaes
30k342101
edited Mar 12 at 20:22
answered Mar 10 at 1:31
ServaesServaes
30k342101
answered Mar 10 at 1:31
ServaesServaes
30k342101
answered Mar 10 at 1:31
ServaesServaes
30k342101
30k342101
$begingroup$
@Glen_b It is reasonable to say has one X to mean there exists an X, though it would be more clear to say there is at least one X.
$endgroup$
– jgon
Mar 12 at 20:21
1
$begingroup$
@Glen_b I have edited to remove the ambiguity.
$endgroup$
– Servaes
Mar 12 at 20:22
add a comment |
$begingroup$
@Glen_b It is reasonable to say has one X to mean there exists an X, though it would be more clear to say there is at least one X.
$endgroup$
– jgon
Mar 12 at 20:21
1
$begingroup$
@Glen_b I have edited to remove the ambiguity.
$endgroup$
– Servaes
Mar 12 at 20:22
$begingroup$
@Glen_b It is reasonable to say has one X to mean there exists an X, though it would be more clear to say there is at least one X.
$endgroup$
– jgon
Mar 12 at 20:21
$begingroup$
@Glen_b It is reasonable to say has one X to mean there exists an X, though it would be more clear to say there is at least one X.
$endgroup$
– jgon
Mar 12 at 20:21
1
1
$begingroup$
@Glen_b I have edited to remove the ambiguity.
$endgroup$
– Servaes
Mar 12 at 20:22
$begingroup$
@Glen_b I have edited to remove the ambiguity.
$endgroup$
– Servaes
Mar 12 at 20:22
add a comment |
$begingroup$
Over the complex numbers, every polynomial factors into roots. So we can take any cubic and write it as $a(x-u)(x-v)(x-w)$ where $u, v, w$ are the roots (they don't need to be distict) and $a$ is the leading coefficient. This lets us form polynomials with only complex roots such as $(x-i)^3$.
However, if all the original coefficients of the polynomial are real, and $c$ is a complex root, then its conjugate $barc$ must also be a root: complex roots to polynomials with real coeffecients must come in pairs. This is called the complex conjugate root theorem.
This means that a polynomial with real coefficients and odd degree will always have at least one real root, which answers the case for cubics. A quadratic with negative discriminant on the other hand has two, conjugate complex roots.
answered Mar 10 at 18:28
BristolBristol
1864
$endgroup$
add a comment |
$begingroup$
Over the complex numbers, every polynomial factors into roots. So we can take any cubic and write it as $a(x-u)(x-v)(x-w)$ where $u, v, w$ are the roots (they don't need to be distict) and $a$ is the leading coefficient. This lets us form polynomials with only complex roots such as $(x-i)^3$.
However, if all the original coefficients of the polynomial are real, and $c$ is a complex root, then its conjugate $barc$ must also be a root: complex roots to polynomials with real coeffecients must come in pairs. This is called the complex conjugate root theorem.
This means that a polynomial with real coefficients and odd degree will always have at least one real root, which answers the case for cubics. A quadratic with negative discriminant on the other hand has two, conjugate complex roots.
answered Mar 10 at 18:28
BristolBristol
1864
$endgroup$
add a comment |
$begingroup$
Over the complex numbers, every polynomial factors into roots. So we can take any cubic and write it as $a(x-u)(x-v)(x-w)$ where $u, v, w$ are the roots (they don't need to be distict) and $a$ is the leading coefficient. This lets us form polynomials with only complex roots such as $(x-i)^3$.
However, if all the original coefficients of the polynomial are real, and $c$ is a complex root, then its conjugate $barc$ must also be a root: complex roots to polynomials with real coeffecients must come in pairs. This is called the complex conjugate root theorem.
This means that a polynomial with real coefficients and odd degree will always have at least one real root, which answers the case for cubics. A quadratic with negative discriminant on the other hand has two, conjugate complex roots.
answered Mar 10 at 18:28
BristolBristol
1864
$endgroup$
Over the complex numbers, every polynomial factors into roots. So we can take any cubic and write it as $a(x-u)(x-v)(x-w)$ where $u, v, w$ are the roots (they don't need to be distict) and $a$ is the leading coefficient. This lets us form polynomials with only complex roots such as $(x-i)^3$.
However, if all the original coefficients of the polynomial are real, and $c$ is a complex root, then its conjugate $barc$ must also be a root: complex roots to polynomials with real coeffecients must come in pairs. This is called the complex conjugate root theorem.
This means that a polynomial with real coefficients and odd degree will always have at least one real root, which answers the case for cubics. A quadratic with negative discriminant on the other hand has two, conjugate complex roots.
answered Mar 10 at 18:28
BristolBristol
1864
edited Mar 10 at 18:43
answered Mar 10 at 18:28
BristolBristol
1864
answered Mar 10 at 18:28
BristolBristol
1864
answered Mar 10 at 18:28
BristolBristol
1864
1864
add a comment |
add a comment |
5
$begingroup$
It's more clear to say "at least one real root" or "a real root". There are real cubics that have three real roots (there are none that have exactly two though).
$endgroup$
– quid♦
Mar 10 at 14:47
6
$begingroup$
@quid: If you count with multiplicity, that's true. If you don't, then you can have $f(x) = (x-1)(x-2)^2$.
$endgroup$
– Kevin
Mar 10 at 17:22
1
$begingroup$
Indeed @Kevin I should have made that explicity.
$endgroup$
– quid♦
Mar 10 at 19:15