Mixing
The sum of three numbers is 0 and their product is 1. Find the sum of the product of any two numbers taken at a time. [closed]
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I am having difficulty wrapping my head around the question in the first place, let alone understanding where I am going wrong. Any help would be appreciated.
algebra-precalculus
asked Jul 14 at 14:30
Satyajit Sen
141
closed as off-topic by Alex Francisco, user223391, steven gregory, Jyrki Lahtonen, Shaun Jul 14 at 17:21
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Community, steven gregory, Jyrki Lahtonen, Shaun
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up vote
0
down vote
favorite
I am having difficulty wrapping my head around the question in the first place, let alone understanding where I am going wrong. Any help would be appreciated.
algebra-precalculus
asked Jul 14 at 14:30
Satyajit Sen
141
closed as off-topic by Alex Francisco, user223391, steven gregory, Jyrki Lahtonen, Shaun Jul 14 at 17:21
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Community, steven gregory, Jyrki Lahtonen, Shaun
There is not enough information to find the sum. It could be any number.
– Somos
Jul 14 at 15:14
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am having difficulty wrapping my head around the question in the first place, let alone understanding where I am going wrong. Any help would be appreciated.
algebra-precalculus
asked Jul 14 at 14:30
Satyajit Sen
141
I am having difficulty wrapping my head around the question in the first place, let alone understanding where I am going wrong. Any help would be appreciated.
algebra-precalculus
asked Jul 14 at 14:30
Satyajit Sen
141
asked Jul 14 at 14:30
Satyajit Sen
141
asked Jul 14 at 14:30
Satyajit Sen
141
asked Jul 14 at 14:30
Satyajit Sen
141
141
closed as off-topic by Alex Francisco, user223391, steven gregory, Jyrki Lahtonen, Shaun Jul 14 at 17:21
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Community, steven gregory, Jyrki Lahtonen, Shaun
closed as off-topic by Alex Francisco, user223391, steven gregory, Jyrki Lahtonen, Shaun Jul 14 at 17:21
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Community, steven gregory, Jyrki Lahtonen, Shaun
There is not enough information to find the sum. It could be any number.
– Somos
Jul 14 at 15:14
add a comment |Â
There is not enough information to find the sum. It could be any number.
– Somos
Jul 14 at 15:14
There is not enough information to find the sum. It could be any number.
– Somos
Jul 14 at 15:14
There is not enough information to find the sum. It could be any number.
– Somos
Jul 14 at 15:14
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
Assuming the numbers $a,b,cinmathbbR$,
we have a cubic equation
beginalign
(x-a)(x-b)(x-c)&=0
,\
x^3-(a+b+c)x^2+(ab+bc+ca)x-abc
&=0
,\
x^3+(ab+bc+ca)x-1
&=0
,\
x^3+ux-1
&=0
,quad
u=ab+bc+ca
tag1label1
.
endalign
The discriminant
of the cubic is
beginalign
Delta&=-27-4u^3
,\
endalign
a condition for three real roots is
beginalign
Delta&ge 0
,\
-27-4u^3&ge 0
,\
4u^3&le-27
,\
u&le-tfrac32cdotsqrt[3]2
approx -1.88988
.
endalign
Thus, for any $u=ab+bc+cale-tfrac32cdotsqrt[3]2$
we can always find a set of three real numbers $a,b,c$
as roots of eqref1, for which $a+b+c=0$ and $abc=1$.
Two arbitrary examples:
beginalign
u&=-2quadmapstoquad a=-1,
quad b=tfrac12+tfracsqrt52,
quad c=tfrac12-tfracsqrt52
;\
u&=-tfrac72quadmapstoquad a=2,
quad b=-1+tfracsqrt22,
quad c=-1-tfracsqrt22
.
endalign
answered Jul 14 at 15:27
g.kov
5,5521717
Can you please provide the answer in case complex roots are to be considered as well?
– Satyajit Sen
Jul 14 at 16:03
@Satyajit Sen: Without limitation that all $a,b,c$ are real, $u$ can be any real number. That is, if we choose some $u>-tfrac32cdotsqrt[3]2$, then just one root of eqref1 would be real, and the other two would be complex conjugate.
– g.kov
Jul 14 at 16:21
@Satyajit Sen: For example, for $u=tfrac74$ we have $a=tfrac12$, $b=-tfrac14+tfracsqrt314,i$, $c=-tfrac14-tfracsqrt314,i$.
– g.kov
Jul 14 at 17:37
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Assuming the numbers $a,b,cinmathbbR$,
we have a cubic equation
beginalign
(x-a)(x-b)(x-c)&=0
,\
x^3-(a+b+c)x^2+(ab+bc+ca)x-abc
&=0
,\
x^3+(ab+bc+ca)x-1
&=0
,\
x^3+ux-1
&=0
,quad
u=ab+bc+ca
tag1label1
.
endalign
The discriminant
of the cubic is
beginalign
Delta&=-27-4u^3
,\
endalign
a condition for three real roots is
beginalign
Delta&ge 0
,\
-27-4u^3&ge 0
,\
4u^3&le-27
,\
u&le-tfrac32cdotsqrt[3]2
approx -1.88988
.
endalign
Thus, for any $u=ab+bc+cale-tfrac32cdotsqrt[3]2$
we can always find a set of three real numbers $a,b,c$
as roots of eqref1, for which $a+b+c=0$ and $abc=1$.
Two arbitrary examples:
beginalign
u&=-2quadmapstoquad a=-1,
quad b=tfrac12+tfracsqrt52,
quad c=tfrac12-tfracsqrt52
;\
u&=-tfrac72quadmapstoquad a=2,
quad b=-1+tfracsqrt22,
quad c=-1-tfracsqrt22
.
endalign
answered Jul 14 at 15:27
g.kov
5,5521717
Can you please provide the answer in case complex roots are to be considered as well?
– Satyajit Sen
Jul 14 at 16:03
@Satyajit Sen: Without limitation that all $a,b,c$ are real, $u$ can be any real number. That is, if we choose some $u>-tfrac32cdotsqrt[3]2$, then just one root of eqref1 would be real, and the other two would be complex conjugate.
– g.kov
Jul 14 at 16:21
@Satyajit Sen: For example, for $u=tfrac74$ we have $a=tfrac12$, $b=-tfrac14+tfracsqrt314,i$, $c=-tfrac14-tfracsqrt314,i$.
– g.kov
Jul 14 at 17:37
add a comment |Â
up vote
1
down vote
Assuming the numbers $a,b,cinmathbbR$,
we have a cubic equation
beginalign
(x-a)(x-b)(x-c)&=0
,\
x^3-(a+b+c)x^2+(ab+bc+ca)x-abc
&=0
,\
x^3+(ab+bc+ca)x-1
&=0
,\
x^3+ux-1
&=0
,quad
u=ab+bc+ca
tag1label1
.
endalign
The discriminant
of the cubic is
beginalign
Delta&=-27-4u^3
,\
endalign
a condition for three real roots is
beginalign
Delta&ge 0
,\
-27-4u^3&ge 0
,\
4u^3&le-27
,\
u&le-tfrac32cdotsqrt[3]2
approx -1.88988
.
endalign
Thus, for any $u=ab+bc+cale-tfrac32cdotsqrt[3]2$
we can always find a set of three real numbers $a,b,c$
as roots of eqref1, for which $a+b+c=0$ and $abc=1$.
Two arbitrary examples:
beginalign
u&=-2quadmapstoquad a=-1,
quad b=tfrac12+tfracsqrt52,
quad c=tfrac12-tfracsqrt52
;\
u&=-tfrac72quadmapstoquad a=2,
quad b=-1+tfracsqrt22,
quad c=-1-tfracsqrt22
.
endalign
answered Jul 14 at 15:27
g.kov
5,5521717
Can you please provide the answer in case complex roots are to be considered as well?
– Satyajit Sen
Jul 14 at 16:03
@Satyajit Sen: Without limitation that all $a,b,c$ are real, $u$ can be any real number. That is, if we choose some $u>-tfrac32cdotsqrt[3]2$, then just one root of eqref1 would be real, and the other two would be complex conjugate.
– g.kov
Jul 14 at 16:21
@Satyajit Sen: For example, for $u=tfrac74$ we have $a=tfrac12$, $b=-tfrac14+tfracsqrt314,i$, $c=-tfrac14-tfracsqrt314,i$.
– g.kov
Jul 14 at 17:37
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Assuming the numbers $a,b,cinmathbbR$,
we have a cubic equation
beginalign
(x-a)(x-b)(x-c)&=0
,\
x^3-(a+b+c)x^2+(ab+bc+ca)x-abc
&=0
,\
x^3+(ab+bc+ca)x-1
&=0
,\
x^3+ux-1
&=0
,quad
u=ab+bc+ca
tag1label1
.
endalign
The discriminant
of the cubic is
beginalign
Delta&=-27-4u^3
,\
endalign
a condition for three real roots is
beginalign
Delta&ge 0
,\
-27-4u^3&ge 0
,\
4u^3&le-27
,\
u&le-tfrac32cdotsqrt[3]2
approx -1.88988
.
endalign
Thus, for any $u=ab+bc+cale-tfrac32cdotsqrt[3]2$
we can always find a set of three real numbers $a,b,c$
as roots of eqref1, for which $a+b+c=0$ and $abc=1$.
Two arbitrary examples:
beginalign
u&=-2quadmapstoquad a=-1,
quad b=tfrac12+tfracsqrt52,
quad c=tfrac12-tfracsqrt52
;\
u&=-tfrac72quadmapstoquad a=2,
quad b=-1+tfracsqrt22,
quad c=-1-tfracsqrt22
.
endalign
answered Jul 14 at 15:27
g.kov
5,5521717
Assuming the numbers $a,b,cinmathbbR$,
we have a cubic equation
beginalign
(x-a)(x-b)(x-c)&=0
,\
x^3-(a+b+c)x^2+(ab+bc+ca)x-abc
&=0
,\
x^3+(ab+bc+ca)x-1
&=0
,\
x^3+ux-1
&=0
,quad
u=ab+bc+ca
tag1label1
.
endalign
The discriminant
of the cubic is
beginalign
Delta&=-27-4u^3
,\
endalign
a condition for three real roots is
beginalign
Delta&ge 0
,\
-27-4u^3&ge 0
,\
4u^3&le-27
,\
u&le-tfrac32cdotsqrt[3]2
approx -1.88988
.
endalign
Thus, for any $u=ab+bc+cale-tfrac32cdotsqrt[3]2$
we can always find a set of three real numbers $a,b,c$
as roots of eqref1, for which $a+b+c=0$ and $abc=1$.
Two arbitrary examples:
beginalign
u&=-2quadmapstoquad a=-1,
quad b=tfrac12+tfracsqrt52,
quad c=tfrac12-tfracsqrt52
;\
u&=-tfrac72quadmapstoquad a=2,
quad b=-1+tfracsqrt22,
quad c=-1-tfracsqrt22
.
endalign
answered Jul 14 at 15:27
g.kov
5,5521717
answered Jul 14 at 15:27
g.kov
5,5521717
answered Jul 14 at 15:27
g.kov
5,5521717
answered Jul 14 at 15:27
g.kov
5,5521717
5,5521717
Can you please provide the answer in case complex roots are to be considered as well?
– Satyajit Sen
Jul 14 at 16:03
@Satyajit Sen: Without limitation that all $a,b,c$ are real, $u$ can be any real number. That is, if we choose some $u>-tfrac32cdotsqrt[3]2$, then just one root of eqref1 would be real, and the other two would be complex conjugate.
– g.kov
Jul 14 at 16:21
@Satyajit Sen: For example, for $u=tfrac74$ we have $a=tfrac12$, $b=-tfrac14+tfracsqrt314,i$, $c=-tfrac14-tfracsqrt314,i$.
– g.kov
Jul 14 at 17:37
add a comment |Â
Can you please provide the answer in case complex roots are to be considered as well?
– Satyajit Sen
Jul 14 at 16:03
@Satyajit Sen: Without limitation that all $a,b,c$ are real, $u$ can be any real number. That is, if we choose some $u>-tfrac32cdotsqrt[3]2$, then just one root of eqref1 would be real, and the other two would be complex conjugate.
– g.kov
Jul 14 at 16:21
@Satyajit Sen: For example, for $u=tfrac74$ we have $a=tfrac12$, $b=-tfrac14+tfracsqrt314,i$, $c=-tfrac14-tfracsqrt314,i$.
– g.kov
Jul 14 at 17:37
Can you please provide the answer in case complex roots are to be considered as well?
– Satyajit Sen
Jul 14 at 16:03
Can you please provide the answer in case complex roots are to be considered as well?
– Satyajit Sen
Jul 14 at 16:03
@Satyajit Sen: Without limitation that all $a,b,c$ are real, $u$ can be any real number. That is, if we choose some $u>-tfrac32cdotsqrt[3]2$, then just one root of eqref1 would be real, and the other two would be complex conjugate.
– g.kov
Jul 14 at 16:21
@Satyajit Sen: Without limitation that all $a,b,c$ are real, $u$ can be any real number. That is, if we choose some $u>-tfrac32cdotsqrt[3]2$, then just one root of eqref1 would be real, and the other two would be complex conjugate.
– g.kov
Jul 14 at 16:21
@Satyajit Sen: For example, for $u=tfrac74$ we have $a=tfrac12$, $b=-tfrac14+tfracsqrt314,i$, $c=-tfrac14-tfracsqrt314,i$.
– g.kov
Jul 14 at 17:37
@Satyajit Sen: For example, for $u=tfrac74$ we have $a=tfrac12$, $b=-tfrac14+tfracsqrt314,i$, $c=-tfrac14-tfracsqrt314,i$.
– g.kov
Jul 14 at 17:37
add a comment |Â
There is not enough information to find the sum. It could be any number.
– Somos
Jul 14 at 15:14