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Proving roots of a function cannot all be real

Seeing complex roots on the graph of a polynomialDifferentiating $ arctan (fraca+x1-ax)- arctan x$Vertical length of two rootsProving roots of a polynomial are real and distinct.What is the maximum number of real roots a polynomial of any degree can have?Polynomials and Location of Real RootsNumber of distinct real roots of derivative of a functionHow is the average of the real and complex roots of a polynomial connected to the average of the real and complex roots of its derivative?

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7

4

$begingroup$

The question is

Let $a$$,b$$,c$$,d$ be any four real number not all equal to zero. Prove that the roots of the polynomial $f(x)=x^6+ax^3+bx^2+cx+d$ cannot all be real.

I know that the first derivative of a functions gives its turning points and the second derivative give its concavity but how does it help to solve the question. Or if you have any other methods it would really help.Any hint would also be appreciated . Thanks in advance.

real-analysis derivatives polynomials graphing-functions

share|cite|improve this question

asked Oct 1 at 15:17

Akshaj BansalAkshaj Bansal

73011 silver badge88 bronze badges

$endgroup$

add a comment
 | 

7

4

$begingroup$

The question is

Let $a$$,b$$,c$$,d$ be any four real number not all equal to zero. Prove that the roots of the polynomial $f(x)=x^6+ax^3+bx^2+cx+d$ cannot all be real.

I know that the first derivative of a functions gives its turning points and the second derivative give its concavity but how does it help to solve the question. Or if you have any other methods it would really help.Any hint would also be appreciated . Thanks in advance.

real-analysis derivatives polynomials graphing-functions

share|cite|improve this question

asked Oct 1 at 15:17

Akshaj BansalAkshaj Bansal

73011 silver badge88 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

The question is

Let $a$$,b$$,c$$,d$ be any four real number not all equal to zero. Prove that the roots of the polynomial $f(x)=x^6+ax^3+bx^2+cx+d$ cannot all be real.

I know that the first derivative of a functions gives its turning points and the second derivative give its concavity but how does it help to solve the question. Or if you have any other methods it would really help.Any hint would also be appreciated . Thanks in advance.

real-analysis derivatives polynomials graphing-functions

share|cite|improve this question

asked Oct 1 at 15:17

Akshaj BansalAkshaj Bansal

73011 silver badge88 bronze badges

$endgroup$

share|cite|improve this question

asked Oct 1 at 15:17

Akshaj BansalAkshaj Bansal

73011 silver badge88 bronze badges

share|cite|improve this question

asked Oct 1 at 15:17

Akshaj BansalAkshaj Bansal

73011 silver badge88 bronze badges

asked Oct 1 at 15:17

Akshaj BansalAkshaj Bansal

73011 silver badge88 bronze badges

asked Oct 1 at 15:17

Akshaj BansalAkshaj Bansal

73011 silver badge88 bronze badges

3 Answers
3

active

oldest

votes

14

$begingroup$

If $f$ has $6$ real roots (perhaps with multiplicity), then $f'(x)$ has $5$ real roots, $f''(x)$ has $4$ real roots, and so on.
We have
$$
beginalign
f(x)&=x^6+ax^3+bx^2+cx+d\
f'(x)&=6x^5+3ax^2+2bx+c\
f''(x)&=30x^4+6ax+2b\
f'''(x)&=120x^3+6a\
endalign $$
Note that $x^m+p$ with $m>2$ has $m$ real roots if and only if $p=0$ (because it can have at most two distinct real roots, and a multiple root only if $p=0$).
Using this, we see from the expression for $f'''(x)$ that $a=0$. Then by the same argument using $f''(x)$, we have $b=0$. After that using $f'(x)$, we obtain $c=0$, and finally using $f(x)$, we find $d=0$. But $a=b=c=d=0$ is excluded in the problem statement.

share|cite|improve this answer

edited Oct 1 at 15:38

answered Oct 1 at 15:29

Hagen von EitzenHagen von Eitzen

305k2525 gold badges291291 silver badges536536 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I am trying to learn. Using the derivative is useful as a test for multiple roots, and in some cases to find a multiple root. In this question, we were not told that there are equal roots. Why did you use the derivative as the base for this answer? Thanks.
  $endgroup$
  – NoChance
  Oct 2 at 3:38
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @NoChance: "If $f(x)$ has $m$ real roots, then $f'(x)$ has $m-1$ real roots" is basically Rolle's theorem plus a little bookkeeping.
  $endgroup$
  – Micah
  Oct 2 at 4:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Micah, thank you for your comment but I am not sure how Rolle's theorem is to do with this. What I don't understand is why did the answer assume that f(x) has repeated roots.
  $endgroup$
  – NoChance
  Oct 2 at 6:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The answer doesn't assume that there are repeated roots; it accommodates the possiblity that there are repeated roots.
  $endgroup$
  – Greg Martin
  Oct 2 at 6:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GregMartin, thanks for your comment.
  $endgroup$
  – NoChance
  Oct 2 at 10:13
  

  
  
  
  

add a comment
 | 

7

$begingroup$

Suppose all the roots of $f(x)$ are real.

By Vieta's formula,

$$textSum of the roots=-sum alpha = 0.$$

Squaring,
$$sum alpha^2 + 2 sum alpha beta = sumalpha^2 + 0 = 0.$$

This is possible only if all the roots are equal to $0$, which is false since $a, b, c, d$ are not all $0$ .

share|cite|improve this answer

edited Oct 2 at 3:12

dmtri

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answered Oct 1 at 15:25

user600016user600016

86611 silver badge1414 bronze badges

$endgroup$

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  $-sumalpha$ is the coefficient of $x^n-1$, and $sumalphabeta$ is the coefficient of $x^n-2$.
  $endgroup$
  – Hagen von Eitzen
  Oct 1 at 15:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Fixed thank you
  $endgroup$
  – user600016
  Oct 1 at 16:08
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  -1, this answer is too confusing for a non-expert. If it is improved with a prelude that explains the notation and/or theorems being used I'll retract my downvote.
  $endgroup$
  – goblin
  Oct 2 at 2:41
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  So please add that information and a link in the answer for those that do not know what you are thinking.
  $endgroup$
  – StephenG
  Oct 2 at 2:58
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Im more curious about the "squaring" comment. Squaring what? Where does beta come from? No explanatory content whatsoever.
  $endgroup$
  – SquishyRhode
  Oct 2 at 3:18
  

  
  
  
  

 | 
show 12 more comments

0

$begingroup$

Descartes rule of signs is independent of the degree sequence, it only demands that the coefficients under consideration are degree-ordered when computing the sign variations. This means that the given polynomial has the same maximal amount of positive, negative and zero roots as $0=x^4+ax^3+bx^2+cx+d$.

Which means that among the roots of the original degree-6 polynomial there are at most $4$ real roots, at least $2$ roots have to be non-real complex.

share|cite|improve this answer

edited Oct 2 at 10:20

answered Oct 2 at 10:01

Dr. Lutz LehmannDr. Lutz Lehmann

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14

$begingroup$

If $f$ has $6$ real roots (perhaps with multiplicity), then $f'(x)$ has $5$ real roots, $f''(x)$ has $4$ real roots, and so on.
We have
$$
beginalign
f(x)&=x^6+ax^3+bx^2+cx+d\
f'(x)&=6x^5+3ax^2+2bx+c\
f''(x)&=30x^4+6ax+2b\
f'''(x)&=120x^3+6a\
endalign $$
Note that $x^m+p$ with $m>2$ has $m$ real roots if and only if $p=0$ (because it can have at most two distinct real roots, and a multiple root only if $p=0$).
Using this, we see from the expression for $f'''(x)$ that $a=0$. Then by the same argument using $f''(x)$, we have $b=0$. After that using $f'(x)$, we obtain $c=0$, and finally using $f(x)$, we find $d=0$. But $a=b=c=d=0$ is excluded in the problem statement.

share|cite|improve this answer

edited Oct 1 at 15:38

answered Oct 1 at 15:29

Hagen von EitzenHagen von Eitzen

305k2525 gold badges291291 silver badges536536 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I am trying to learn. Using the derivative is useful as a test for multiple roots, and in some cases to find a multiple root. In this question, we were not told that there are equal roots. Why did you use the derivative as the base for this answer? Thanks.
  $endgroup$
  – NoChance
  Oct 2 at 3:38
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @NoChance: "If $f(x)$ has $m$ real roots, then $f'(x)$ has $m-1$ real roots" is basically Rolle's theorem plus a little bookkeeping.
  $endgroup$
  – Micah
  Oct 2 at 4:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Micah, thank you for your comment but I am not sure how Rolle's theorem is to do with this. What I don't understand is why did the answer assume that f(x) has repeated roots.
  $endgroup$
  – NoChance
  Oct 2 at 6:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The answer doesn't assume that there are repeated roots; it accommodates the possiblity that there are repeated roots.
  $endgroup$
  – Greg Martin
  Oct 2 at 6:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GregMartin, thanks for your comment.
  $endgroup$
  – NoChance
  Oct 2 at 10:13
  

  
  
  
  

add a comment
 | 

14

$begingroup$

If $f$ has $6$ real roots (perhaps with multiplicity), then $f'(x)$ has $5$ real roots, $f''(x)$ has $4$ real roots, and so on.
We have
$$
beginalign
f(x)&=x^6+ax^3+bx^2+cx+d\
f'(x)&=6x^5+3ax^2+2bx+c\
f''(x)&=30x^4+6ax+2b\
f'''(x)&=120x^3+6a\
endalign $$
Note that $x^m+p$ with $m>2$ has $m$ real roots if and only if $p=0$ (because it can have at most two distinct real roots, and a multiple root only if $p=0$).
Using this, we see from the expression for $f'''(x)$ that $a=0$. Then by the same argument using $f''(x)$, we have $b=0$. After that using $f'(x)$, we obtain $c=0$, and finally using $f(x)$, we find $d=0$. But $a=b=c=d=0$ is excluded in the problem statement.

share|cite|improve this answer

edited Oct 1 at 15:38

answered Oct 1 at 15:29

Hagen von EitzenHagen von Eitzen

305k2525 gold badges291291 silver badges536536 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I am trying to learn. Using the derivative is useful as a test for multiple roots, and in some cases to find a multiple root. In this question, we were not told that there are equal roots. Why did you use the derivative as the base for this answer? Thanks.
  $endgroup$
  – NoChance
  Oct 2 at 3:38
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @NoChance: "If $f(x)$ has $m$ real roots, then $f'(x)$ has $m-1$ real roots" is basically Rolle's theorem plus a little bookkeeping.
  $endgroup$
  – Micah
  Oct 2 at 4:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Micah, thank you for your comment but I am not sure how Rolle's theorem is to do with this. What I don't understand is why did the answer assume that f(x) has repeated roots.
  $endgroup$
  – NoChance
  Oct 2 at 6:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The answer doesn't assume that there are repeated roots; it accommodates the possiblity that there are repeated roots.
  $endgroup$
  – Greg Martin
  Oct 2 at 6:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GregMartin, thanks for your comment.
  $endgroup$
  – NoChance
  Oct 2 at 10:13
  

  
  
  
  

add a comment
 | 

$begingroup$

If $f$ has $6$ real roots (perhaps with multiplicity), then $f'(x)$ has $5$ real roots, $f''(x)$ has $4$ real roots, and so on.
We have
$$
beginalign
f(x)&=x^6+ax^3+bx^2+cx+d\
f'(x)&=6x^5+3ax^2+2bx+c\
f''(x)&=30x^4+6ax+2b\
f'''(x)&=120x^3+6a\
endalign $$
Note that $x^m+p$ with $m>2$ has $m$ real roots if and only if $p=0$ (because it can have at most two distinct real roots, and a multiple root only if $p=0$).
Using this, we see from the expression for $f'''(x)$ that $a=0$. Then by the same argument using $f''(x)$, we have $b=0$. After that using $f'(x)$, we obtain $c=0$, and finally using $f(x)$, we find $d=0$. But $a=b=c=d=0$ is excluded in the problem statement.

share|cite|improve this answer

edited Oct 1 at 15:38

answered Oct 1 at 15:29

Hagen von EitzenHagen von Eitzen

305k2525 gold badges291291 silver badges536536 bronze badges

$endgroup$

share|cite|improve this answer

edited Oct 1 at 15:38

answered Oct 1 at 15:29

Hagen von EitzenHagen von Eitzen

305k2525 gold badges291291 silver badges536536 bronze badges

answered Oct 1 at 15:29

Hagen von EitzenHagen von Eitzen

305k2525 gold badges291291 silver badges536536 bronze badges

answered Oct 1 at 15:29

Hagen von EitzenHagen von Eitzen

305k2525 gold badges291291 silver badges536536 bronze badges

7

$begingroup$

Suppose all the roots of $f(x)$ are real.

By Vieta's formula,

$$textSum of the roots=-sum alpha = 0.$$

Squaring,
$$sum alpha^2 + 2 sum alpha beta = sumalpha^2 + 0 = 0.$$

This is possible only if all the roots are equal to $0$, which is false since $a, b, c, d$ are not all $0$ .

share|cite|improve this answer

edited Oct 2 at 3:12

dmtri

2,30533 gold badges66 silver badges2323 bronze badges

answered Oct 1 at 15:25

user600016user600016

86611 silver badge1414 bronze badges

$endgroup$

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  $-sumalpha$ is the coefficient of $x^n-1$, and $sumalphabeta$ is the coefficient of $x^n-2$.
  $endgroup$
  – Hagen von Eitzen
  Oct 1 at 15:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Fixed thank you
  $endgroup$
  – user600016
  Oct 1 at 16:08
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  -1, this answer is too confusing for a non-expert. If it is improved with a prelude that explains the notation and/or theorems being used I'll retract my downvote.
  $endgroup$
  – goblin
  Oct 2 at 2:41
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  So please add that information and a link in the answer for those that do not know what you are thinking.
  $endgroup$
  – StephenG
  Oct 2 at 2:58
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Im more curious about the "squaring" comment. Squaring what? Where does beta come from? No explanatory content whatsoever.
  $endgroup$
  – SquishyRhode
  Oct 2 at 3:18
  

  
  
  
  

 | 
show 12 more comments

7

$begingroup$

Suppose all the roots of $f(x)$ are real.

By Vieta's formula,

$$textSum of the roots=-sum alpha = 0.$$

Squaring,
$$sum alpha^2 + 2 sum alpha beta = sumalpha^2 + 0 = 0.$$

This is possible only if all the roots are equal to $0$, which is false since $a, b, c, d$ are not all $0$ .

share|cite|improve this answer

edited Oct 2 at 3:12

dmtri

2,30533 gold badges66 silver badges2323 bronze badges

answered Oct 1 at 15:25

user600016user600016

86611 silver badge1414 bronze badges

$endgroup$

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  $-sumalpha$ is the coefficient of $x^n-1$, and $sumalphabeta$ is the coefficient of $x^n-2$.
  $endgroup$
  – Hagen von Eitzen
  Oct 1 at 15:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Fixed thank you
  $endgroup$
  – user600016
  Oct 1 at 16:08
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  -1, this answer is too confusing for a non-expert. If it is improved with a prelude that explains the notation and/or theorems being used I'll retract my downvote.
  $endgroup$
  – goblin
  Oct 2 at 2:41
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  So please add that information and a link in the answer for those that do not know what you are thinking.
  $endgroup$
  – StephenG
  Oct 2 at 2:58
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Im more curious about the "squaring" comment. Squaring what? Where does beta come from? No explanatory content whatsoever.
  $endgroup$
  – SquishyRhode
  Oct 2 at 3:18
  

  
  
  
  

 | 
show 12 more comments

$begingroup$

Suppose all the roots of $f(x)$ are real.

By Vieta's formula,

$$textSum of the roots=-sum alpha = 0.$$

Squaring,
$$sum alpha^2 + 2 sum alpha beta = sumalpha^2 + 0 = 0.$$

This is possible only if all the roots are equal to $0$, which is false since $a, b, c, d$ are not all $0$ .

share|cite|improve this answer

edited Oct 2 at 3:12

dmtri

2,30533 gold badges66 silver badges2323 bronze badges

answered Oct 1 at 15:25

user600016user600016

86611 silver badge1414 bronze badges

$endgroup$

share|cite|improve this answer

edited Oct 2 at 3:12

dmtri

2,30533 gold badges66 silver badges2323 bronze badges

answered Oct 1 at 15:25

user600016user600016

86611 silver badge1414 bronze badges

edited Oct 2 at 3:12

dmtri

2,30533 gold badges66 silver badges2323 bronze badges

edited Oct 2 at 3:12

dmtri

2,30533 gold badges66 silver badges2323 bronze badges

answered Oct 1 at 15:25

user600016user600016

86611 silver badge1414 bronze badges

answered Oct 1 at 15:25

user600016user600016

86611 silver badge1414 bronze badges

0

$begingroup$

Descartes rule of signs is independent of the degree sequence, it only demands that the coefficients under consideration are degree-ordered when computing the sign variations. This means that the given polynomial has the same maximal amount of positive, negative and zero roots as $0=x^4+ax^3+bx^2+cx+d$.

Which means that among the roots of the original degree-6 polynomial there are at most $4$ real roots, at least $2$ roots have to be non-real complex.

share|cite|improve this answer

edited Oct 2 at 10:20

answered Oct 2 at 10:01

Dr. Lutz LehmannDr. Lutz Lehmann

75.9k55 gold badges2424 silver badges6969 bronze badges

$endgroup$

add a comment
 | 

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Descartes rule of signs is independent of the degree sequence, it only demands that the coefficients under consideration are degree-ordered when computing the sign variations. This means that the given polynomial has the same maximal amount of positive, negative and zero roots as $0=x^4+ax^3+bx^2+cx+d$.

Which means that among the roots of the original degree-6 polynomial there are at most $4$ real roots, at least $2$ roots have to be non-real complex.

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edited Oct 2 at 10:20

answered Oct 2 at 10:01

Dr. Lutz LehmannDr. Lutz Lehmann

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Descartes rule of signs is independent of the degree sequence, it only demands that the coefficients under consideration are degree-ordered when computing the sign variations. This means that the given polynomial has the same maximal amount of positive, negative and zero roots as $0=x^4+ax^3+bx^2+cx+d$.

Which means that among the roots of the original degree-6 polynomial there are at most $4$ real roots, at least $2$ roots have to be non-real complex.

share|cite|improve this answer

edited Oct 2 at 10:20

answered Oct 2 at 10:01

Dr. Lutz LehmannDr. Lutz Lehmann

75.9k55 gold badges2424 silver badges6969 bronze badges

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share|cite|improve this answer

edited Oct 2 at 10:20

answered Oct 2 at 10:01

Dr. Lutz LehmannDr. Lutz Lehmann

75.9k55 gold badges2424 silver badges6969 bronze badges

answered Oct 2 at 10:01

Dr. Lutz LehmannDr. Lutz Lehmann

75.9k55 gold badges2424 silver badges6969 bronze badges

answered Oct 2 at 10:01

Dr. Lutz LehmannDr. Lutz Lehmann

75.9k55 gold badges2424 silver badges6969 bronze badges