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Given that $f(x) :=x^6-6x^5+ax^4+bx^3+cx^2+dx+1$ has its roots as all positive, find $a,b,c,d$
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From Larson's "Problem Solving Through Problems" 7.2.10:
Given that $f(x) :=x^6-6x^5+ax^4+bx^3+cx^2+dx+1$ has its roots as all positive, find $a,b,c,d$
Thus chapter is about (generalized) Arithmetic-geometric mean inequality so I would have to use that. I believe it's implied from the problem that all roots are real.
Any hints or solution would be great.
problem-solving
asked Jul 20 at 18:42
BlueRoses
838314
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up vote
7
down vote
favorite
From Larson's "Problem Solving Through Problems" 7.2.10:
Given that $f(x) :=x^6-6x^5+ax^4+bx^3+cx^2+dx+1$ has its roots as all positive, find $a,b,c,d$
Thus chapter is about (generalized) Arithmetic-geometric mean inequality so I would have to use that. I believe it's implied from the problem that all roots are real.
Any hints or solution would be great.
problem-solving
asked Jul 20 at 18:42
BlueRoses
838314
11
$6$ is the sum of the roots and $1$ the product. Look at the condition when the arithmetic mean is equal to the geometric mean. Not just knowing that the roots are real, but positive.
– user577471
Jul 20 at 18:44
7
I got it, since the sum of roots divided by $6$ is actually equal to the geometric mean of the roots, it must be that they are all equal to $1$ and hence we just have to expand $(x-1)^6$
– BlueRoses
Jul 20 at 18:49
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
From Larson's "Problem Solving Through Problems" 7.2.10:
Given that $f(x) :=x^6-6x^5+ax^4+bx^3+cx^2+dx+1$ has its roots as all positive, find $a,b,c,d$
Thus chapter is about (generalized) Arithmetic-geometric mean inequality so I would have to use that. I believe it's implied from the problem that all roots are real.
Any hints or solution would be great.
problem-solving
asked Jul 20 at 18:42
BlueRoses
838314
From Larson's "Problem Solving Through Problems" 7.2.10:
Given that $f(x) :=x^6-6x^5+ax^4+bx^3+cx^2+dx+1$ has its roots as all positive, find $a,b,c,d$
Thus chapter is about (generalized) Arithmetic-geometric mean inequality so I would have to use that. I believe it's implied from the problem that all roots are real.
Any hints or solution would be great.
problem-solving
asked Jul 20 at 18:42
BlueRoses
838314
asked Jul 20 at 18:42
BlueRoses
838314
asked Jul 20 at 18:42
BlueRoses
838314
asked Jul 20 at 18:42
BlueRoses
838314
838314
11
$6$ is the sum of the roots and $1$ the product. Look at the condition when the arithmetic mean is equal to the geometric mean. Not just knowing that the roots are real, but positive.
– user577471
Jul 20 at 18:44
7
I got it, since the sum of roots divided by $6$ is actually equal to the geometric mean of the roots, it must be that they are all equal to $1$ and hence we just have to expand $(x-1)^6$
– BlueRoses
Jul 20 at 18:49
add a comment |Â
11
$6$ is the sum of the roots and $1$ the product. Look at the condition when the arithmetic mean is equal to the geometric mean. Not just knowing that the roots are real, but positive.
– user577471
Jul 20 at 18:44
7
I got it, since the sum of roots divided by $6$ is actually equal to the geometric mean of the roots, it must be that they are all equal to $1$ and hence we just have to expand $(x-1)^6$
– BlueRoses
Jul 20 at 18:49
11
11
$6$ is the sum of the roots and $1$ the product. Look at the condition when the arithmetic mean is equal to the geometric mean. Not just knowing that the roots are real, but positive.
– user577471
Jul 20 at 18:44
$6$ is the sum of the roots and $1$ the product. Look at the condition when the arithmetic mean is equal to the geometric mean. Not just knowing that the roots are real, but positive.
– user577471
Jul 20 at 18:44
7
7
I got it, since the sum of roots divided by $6$ is actually equal to the geometric mean of the roots, it must be that they are all equal to $1$ and hence we just have to expand $(x-1)^6$
– BlueRoses
Jul 20 at 18:49
I got it, since the sum of roots divided by $6$ is actually equal to the geometric mean of the roots, it must be that they are all equal to $1$ and hence we just have to expand $(x-1)^6$
– BlueRoses
Jul 20 at 18:49
add a comment |Â
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11
$6$ is the sum of the roots and $1$ the product. Look at the condition when the arithmetic mean is equal to the geometric mean. Not just knowing that the roots are real, but positive.
– user577471
Jul 20 at 18:44
7
I got it, since the sum of roots divided by $6$ is actually equal to the geometric mean of the roots, it must be that they are all equal to $1$ and hence we just have to expand $(x-1)^6$
– BlueRoses
Jul 20 at 18:49