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Proving roots of a function cannot all be real

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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$Question about quartic equation having all 4 real rootsVertical 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 functionDetermine the values of k for which the equation will have zero, two or four real roots

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4

1

$begingroup$

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

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asked 8 hours ago

Akshaj BansalAkshaj Bansal

44788 bronze badges

$endgroup$

add a comment
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4

1

$begingroup$

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 8 hours ago

Akshaj BansalAkshaj Bansal

44788 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 8 hours ago

Akshaj BansalAkshaj Bansal

44788 bronze badges

$endgroup$

share|cite|improve this question

asked 8 hours ago

Akshaj BansalAkshaj Bansal

44788 bronze badges

share|cite|improve this question

asked 8 hours ago

Akshaj BansalAkshaj Bansal

44788 bronze badges

asked 8 hours ago

Akshaj BansalAkshaj Bansal

44788 bronze badges

asked 8 hours ago

Akshaj BansalAkshaj Bansal

44788 bronze badges

2 Answers
2

active

oldest

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4

$begingroup$

Suppose all roots are real.

Sum of the roots=-$Sigma alpha = 0$.

Squaring, $Sigma alpha^2 + 2 Sigma alpha beta = Sigma alpha^2 + 0 = 0.$ this is possible only of all the roots are equal to $0$, which is false since all of $a,b,c,d$ are not $0$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

thewitnessthewitness

51611 silver badge1313 bronze badges

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  $-sumalpha$ is the coefficient of $x^n-1$, and $sumalphabeta$ is the coefficient of $x^n-2$.
  $endgroup$
  – Hagen von Eitzen
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Fixed thank you
  $endgroup$
  – thewitness
  7 hours ago
  

  
  
  
  

add a comment
 | 

7

$begingroup$

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 8 hours ago

answered 8 hours ago

Hagen von EitzenHagen von Eitzen

299k2424 gold badges288288 silver badges530530 bronze badges

$endgroup$

add a comment
 | 

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4

$begingroup$

Suppose all roots are real.

Sum of the roots=-$Sigma alpha = 0$.

Squaring, $Sigma alpha^2 + 2 Sigma alpha beta = Sigma alpha^2 + 0 = 0.$ this is possible only of all the roots are equal to $0$, which is false since all of $a,b,c,d$ are not $0$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

thewitnessthewitness

51611 silver badge1313 bronze badges

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  $-sumalpha$ is the coefficient of $x^n-1$, and $sumalphabeta$ is the coefficient of $x^n-2$.
  $endgroup$
  – Hagen von Eitzen
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Fixed thank you
  $endgroup$
  – thewitness
  7 hours ago
  

  
  
  
  

add a comment
 | 

4

$begingroup$

Suppose all roots are real.

Sum of the roots=-$Sigma alpha = 0$.

Squaring, $Sigma alpha^2 + 2 Sigma alpha beta = Sigma alpha^2 + 0 = 0.$ this is possible only of all the roots are equal to $0$, which is false since all of $a,b,c,d$ are not $0$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

thewitnessthewitness

51611 silver badge1313 bronze badges

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  $-sumalpha$ is the coefficient of $x^n-1$, and $sumalphabeta$ is the coefficient of $x^n-2$.
  $endgroup$
  – Hagen von Eitzen
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Fixed thank you
  $endgroup$
  – thewitness
  7 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

Suppose all roots are real.

Sum of the roots=-$Sigma alpha = 0$.

Squaring, $Sigma alpha^2 + 2 Sigma alpha beta = Sigma alpha^2 + 0 = 0.$ this is possible only of all the roots are equal to $0$, which is false since all of $a,b,c,d$ are not $0$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

thewitnessthewitness

51611 silver badge1313 bronze badges

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

thewitnessthewitness

51611 silver badge1313 bronze badges

answered 8 hours ago

thewitnessthewitness

51611 silver badge1313 bronze badges

answered 8 hours ago

thewitnessthewitness

51611 silver badge1313 bronze badges

7

$begingroup$

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 8 hours ago

answered 8 hours ago

Hagen von EitzenHagen von Eitzen

299k2424 gold badges288288 silver badges530530 bronze badges

$endgroup$

add a comment
 | 

7

$begingroup$

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 8 hours ago

answered 8 hours ago

Hagen von EitzenHagen von Eitzen

299k2424 gold badges288288 silver badges530530 bronze badges

$endgroup$

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 8 hours ago

answered 8 hours ago

Hagen von EitzenHagen von Eitzen

299k2424 gold badges288288 silver badges530530 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Hagen von EitzenHagen von Eitzen

299k2424 gold badges288288 silver badges530530 bronze badges

answered 8 hours ago

Hagen von EitzenHagen von Eitzen

299k2424 gold badges288288 silver badges530530 bronze badges

answered 8 hours ago

Hagen von EitzenHagen von Eitzen

299k2424 gold badges288288 silver badges530530 bronze badges