Is $x^3+x=y^3+2y-1$ a function? Why? [closed]
$begingroup$
This equation was given:
$$x^3+2x=y^3+2y-1$$
Now please tell me some hints about how I can check if
$$x_1=x_2$$
Then $$y_1=y_2$$
And which differences of squares are needed to rewrite $y$ in terms of $x$?
algebra-precalculus functions
asked Dec 16 '18 at 16:13
user602338user602338
1737
$endgroup$
closed as unclear what you're asking by Dietrich Burde, Saad, Cesareo, Lee David Chung Lin, mrtaurho Dec 21 '18 at 10:44
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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show 1 more comment
$begingroup$
This equation was given:
$$x^3+2x=y^3+2y-1$$
Now please tell me some hints about how I can check if
$$x_1=x_2$$
Then $$y_1=y_2$$
And which differences of squares are needed to rewrite $y$ in terms of $x$?
algebra-precalculus functions
asked Dec 16 '18 at 16:13
user602338user602338
1737
$endgroup$
closed as unclear what you're asking by Dietrich Burde, Saad, Cesareo, Lee David Chung Lin, mrtaurho Dec 21 '18 at 10:44
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
I don't understand $x_1=x_2$ here. What do you want to ask? If $2=2$ then what?
$endgroup$
– Dietrich Burde
Dec 16 '18 at 16:15
$begingroup$
Oh i m blind! Sorry .edited thnks
$endgroup$
– user602338
Dec 16 '18 at 16:16
$begingroup$
Your question doesn't make much sense. What do you want to know, if you can express $y$ in terms of $x$ as a function ?
$endgroup$
– Rebellos
Dec 16 '18 at 16:17
$begingroup$
Further to @DietrichBurde's question, you have $+x$ in the question but $+2x$ in the body.
$endgroup$
– J.G.
Dec 16 '18 at 16:17
$begingroup$
Check if we have 2 equal x , then y woudl be equal for them
$endgroup$
– user602338
Dec 16 '18 at 16:18
|
show 1 more comment
$begingroup$
This equation was given:
$$x^3+2x=y^3+2y-1$$
Now please tell me some hints about how I can check if
$$x_1=x_2$$
Then $$y_1=y_2$$
And which differences of squares are needed to rewrite $y$ in terms of $x$?
algebra-precalculus functions
asked Dec 16 '18 at 16:13
user602338user602338
1737
$endgroup$
This equation was given:
$$x^3+2x=y^3+2y-1$$
Now please tell me some hints about how I can check if
$$x_1=x_2$$
Then $$y_1=y_2$$
And which differences of squares are needed to rewrite $y$ in terms of $x$?
algebra-precalculus functions
algebra-precalculus functions
asked Dec 16 '18 at 16:13
user602338user602338
1737
asked Dec 16 '18 at 16:13
user602338user602338
1737
edited Dec 16 '18 at 16:20
user602338
asked Dec 16 '18 at 16:13
user602338user602338
1737
asked Dec 16 '18 at 16:13
user602338user602338
1737
asked Dec 16 '18 at 16:13
user602338user602338
1737
1737
closed as unclear what you're asking by Dietrich Burde, Saad, Cesareo, Lee David Chung Lin, mrtaurho Dec 21 '18 at 10:44
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Dietrich Burde, Saad, Cesareo, Lee David Chung Lin, mrtaurho Dec 21 '18 at 10:44
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
I don't understand $x_1=x_2$ here. What do you want to ask? If $2=2$ then what?
$endgroup$
– Dietrich Burde
Dec 16 '18 at 16:15
$begingroup$
Oh i m blind! Sorry .edited thnks
$endgroup$
– user602338
Dec 16 '18 at 16:16
$begingroup$
Your question doesn't make much sense. What do you want to know, if you can express $y$ in terms of $x$ as a function ?
$endgroup$
– Rebellos
Dec 16 '18 at 16:17
$begingroup$
Further to @DietrichBurde's question, you have $+x$ in the question but $+2x$ in the body.
$endgroup$
– J.G.
Dec 16 '18 at 16:17
$begingroup$
Check if we have 2 equal x , then y woudl be equal for them
$endgroup$
– user602338
Dec 16 '18 at 16:18
|
show 1 more comment
$begingroup$
I don't understand $x_1=x_2$ here. What do you want to ask? If $2=2$ then what?
$endgroup$
– Dietrich Burde
Dec 16 '18 at 16:15
$begingroup$
Oh i m blind! Sorry .edited thnks
$endgroup$
– user602338
Dec 16 '18 at 16:16
$begingroup$
Your question doesn't make much sense. What do you want to know, if you can express $y$ in terms of $x$ as a function ?
$endgroup$
– Rebellos
Dec 16 '18 at 16:17
$begingroup$
Further to @DietrichBurde's question, you have $+x$ in the question but $+2x$ in the body.
$endgroup$
– J.G.
Dec 16 '18 at 16:17
$begingroup$
Check if we have 2 equal x , then y woudl be equal for them
$endgroup$
– user602338
Dec 16 '18 at 16:18
$begingroup$
I don't understand $x_1=x_2$ here. What do you want to ask? If $2=2$ then what?
$endgroup$
– Dietrich Burde
Dec 16 '18 at 16:15
$begingroup$
I don't understand $x_1=x_2$ here. What do you want to ask? If $2=2$ then what?
$endgroup$
– Dietrich Burde
Dec 16 '18 at 16:15
$begingroup$
Oh i m blind! Sorry .edited thnks
$endgroup$
– user602338
Dec 16 '18 at 16:16
$begingroup$
Oh i m blind! Sorry .edited thnks
$endgroup$
– user602338
Dec 16 '18 at 16:16
$begingroup$
Your question doesn't make much sense. What do you want to know, if you can express $y$ in terms of $x$ as a function ?
$endgroup$
– Rebellos
Dec 16 '18 at 16:17
$begingroup$
Your question doesn't make much sense. What do you want to know, if you can express $y$ in terms of $x$ as a function ?
$endgroup$
– Rebellos
Dec 16 '18 at 16:17
$begingroup$
Further to @DietrichBurde's question, you have $+x$ in the question but $+2x$ in the body.
$endgroup$
– J.G.
Dec 16 '18 at 16:17
$begingroup$
Further to @DietrichBurde's question, you have $+x$ in the question but $+2x$ in the body.
$endgroup$
– J.G.
Dec 16 '18 at 16:17
$begingroup$
Check if we have 2 equal x , then y woudl be equal for them
$endgroup$
– user602338
Dec 16 '18 at 16:18
$begingroup$
Check if we have 2 equal x , then y woudl be equal for them
$endgroup$
– user602338
Dec 16 '18 at 16:18
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
More precisely, you want to know if, given
begin{cases}
y_1^3+2y_1-1=x^3+x \[4px]
y_2^3+2y_2-1=x^3+x
end{cases}
you can conclude that $y_1=y_2$.
The two equations imply
$$
y_1^3-y_2^3+2y_1-2y_2=0
$$
that factors as
$$
(y_1-y_2)(y_1^2+y_1y_2+y_2^2+2)=0
$$
Can you show that the second factor is positive for any $y_1$ and $y_2$?
answered Dec 16 '18 at 17:08
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
Both sides are strictly increasing cubic functions of their variable - for example, you can show this with calculus - so they each never take the same value twice. This implies $x$ is a function of $y$ and vice versa. But if you really want a formula for the result in either direction, I think you'll need to solve (e.g. by Cardano's method - think of it as cubics' equivalent of the quadratic formula.), and the result won't be neat.
answered Dec 16 '18 at 16:20
J.G.J.G.
29.2k22845
$endgroup$
$begingroup$
I want to write y in terms of x. How do I do that?
$endgroup$
– user602338
Dec 16 '18 at 16:22
$begingroup$
@user602338 It's a way for solving a cubic equation. It's like the quadratic formula for cubic equations.
$endgroup$
– Ovi
Dec 16 '18 at 16:22
$begingroup$
@user602338 You can do that using Cardano's method brilliant.org/wiki/cardano-method
$endgroup$
– Ovi
Dec 16 '18 at 16:23
$begingroup$
@user602338 You talk about completing the square in your question... are you sure it wasn't supposed to be squared instead of cubed?
$endgroup$
– Ovi
Dec 16 '18 at 16:23
$begingroup$
@Ovi Sorry but I really don't know how! Please help me with that!
$endgroup$
– user602338
Dec 16 '18 at 16:43
|
show 2 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
More precisely, you want to know if, given
begin{cases}
y_1^3+2y_1-1=x^3+x \[4px]
y_2^3+2y_2-1=x^3+x
end{cases}
you can conclude that $y_1=y_2$.
The two equations imply
$$
y_1^3-y_2^3+2y_1-2y_2=0
$$
that factors as
$$
(y_1-y_2)(y_1^2+y_1y_2+y_2^2+2)=0
$$
Can you show that the second factor is positive for any $y_1$ and $y_2$?
answered Dec 16 '18 at 17:08
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
More precisely, you want to know if, given
begin{cases}
y_1^3+2y_1-1=x^3+x \[4px]
y_2^3+2y_2-1=x^3+x
end{cases}
you can conclude that $y_1=y_2$.
The two equations imply
$$
y_1^3-y_2^3+2y_1-2y_2=0
$$
that factors as
$$
(y_1-y_2)(y_1^2+y_1y_2+y_2^2+2)=0
$$
Can you show that the second factor is positive for any $y_1$ and $y_2$?
answered Dec 16 '18 at 17:08
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
More precisely, you want to know if, given
begin{cases}
y_1^3+2y_1-1=x^3+x \[4px]
y_2^3+2y_2-1=x^3+x
end{cases}
you can conclude that $y_1=y_2$.
The two equations imply
$$
y_1^3-y_2^3+2y_1-2y_2=0
$$
that factors as
$$
(y_1-y_2)(y_1^2+y_1y_2+y_2^2+2)=0
$$
Can you show that the second factor is positive for any $y_1$ and $y_2$?
answered Dec 16 '18 at 17:08
egregegreg
184k1486205
$endgroup$
More precisely, you want to know if, given
begin{cases}
y_1^3+2y_1-1=x^3+x \[4px]
y_2^3+2y_2-1=x^3+x
end{cases}
you can conclude that $y_1=y_2$.
The two equations imply
$$
y_1^3-y_2^3+2y_1-2y_2=0
$$
that factors as
$$
(y_1-y_2)(y_1^2+y_1y_2+y_2^2+2)=0
$$
Can you show that the second factor is positive for any $y_1$ and $y_2$?
answered Dec 16 '18 at 17:08
egregegreg
184k1486205
answered Dec 16 '18 at 17:08
egregegreg
184k1486205
answered Dec 16 '18 at 17:08
egregegreg
184k1486205
answered Dec 16 '18 at 17:08
egregegreg
184k1486205
184k1486205
add a comment |
add a comment |
$begingroup$
Both sides are strictly increasing cubic functions of their variable - for example, you can show this with calculus - so they each never take the same value twice. This implies $x$ is a function of $y$ and vice versa. But if you really want a formula for the result in either direction, I think you'll need to solve (e.g. by Cardano's method - think of it as cubics' equivalent of the quadratic formula.), and the result won't be neat.
answered Dec 16 '18 at 16:20
J.G.J.G.
29.2k22845
$endgroup$
$begingroup$
I want to write y in terms of x. How do I do that?
$endgroup$
– user602338
Dec 16 '18 at 16:22
$begingroup$
@user602338 It's a way for solving a cubic equation. It's like the quadratic formula for cubic equations.
$endgroup$
– Ovi
Dec 16 '18 at 16:22
$begingroup$
@user602338 You can do that using Cardano's method brilliant.org/wiki/cardano-method
$endgroup$
– Ovi
Dec 16 '18 at 16:23
$begingroup$
@user602338 You talk about completing the square in your question... are you sure it wasn't supposed to be squared instead of cubed?
$endgroup$
– Ovi
Dec 16 '18 at 16:23
$begingroup$
@Ovi Sorry but I really don't know how! Please help me with that!
$endgroup$
– user602338
Dec 16 '18 at 16:43
|
show 2 more comments
$begingroup$
Both sides are strictly increasing cubic functions of their variable - for example, you can show this with calculus - so they each never take the same value twice. This implies $x$ is a function of $y$ and vice versa. But if you really want a formula for the result in either direction, I think you'll need to solve (e.g. by Cardano's method - think of it as cubics' equivalent of the quadratic formula.), and the result won't be neat.
answered Dec 16 '18 at 16:20
J.G.J.G.
29.2k22845
$endgroup$
$begingroup$
I want to write y in terms of x. How do I do that?
$endgroup$
– user602338
Dec 16 '18 at 16:22
$begingroup$
@user602338 It's a way for solving a cubic equation. It's like the quadratic formula for cubic equations.
$endgroup$
– Ovi
Dec 16 '18 at 16:22
$begingroup$
@user602338 You can do that using Cardano's method brilliant.org/wiki/cardano-method
$endgroup$
– Ovi
Dec 16 '18 at 16:23
$begingroup$
@user602338 You talk about completing the square in your question... are you sure it wasn't supposed to be squared instead of cubed?
$endgroup$
– Ovi
Dec 16 '18 at 16:23
$begingroup$
@Ovi Sorry but I really don't know how! Please help me with that!
$endgroup$
– user602338
Dec 16 '18 at 16:43
|
show 2 more comments
$begingroup$
Both sides are strictly increasing cubic functions of their variable - for example, you can show this with calculus - so they each never take the same value twice. This implies $x$ is a function of $y$ and vice versa. But if you really want a formula for the result in either direction, I think you'll need to solve (e.g. by Cardano's method - think of it as cubics' equivalent of the quadratic formula.), and the result won't be neat.
answered Dec 16 '18 at 16:20
J.G.J.G.
29.2k22845
$endgroup$
Both sides are strictly increasing cubic functions of their variable - for example, you can show this with calculus - so they each never take the same value twice. This implies $x$ is a function of $y$ and vice versa. But if you really want a formula for the result in either direction, I think you'll need to solve (e.g. by Cardano's method - think of it as cubics' equivalent of the quadratic formula.), and the result won't be neat.
answered Dec 16 '18 at 16:20
J.G.J.G.
29.2k22845
edited Dec 16 '18 at 16:23
answered Dec 16 '18 at 16:20
J.G.J.G.
29.2k22845
answered Dec 16 '18 at 16:20
J.G.J.G.
29.2k22845
answered Dec 16 '18 at 16:20
J.G.J.G.
29.2k22845
29.2k22845
$begingroup$
I want to write y in terms of x. How do I do that?
$endgroup$
– user602338
Dec 16 '18 at 16:22
$begingroup$
@user602338 It's a way for solving a cubic equation. It's like the quadratic formula for cubic equations.
$endgroup$
– Ovi
Dec 16 '18 at 16:22
$begingroup$
@user602338 You can do that using Cardano's method brilliant.org/wiki/cardano-method
$endgroup$
– Ovi
Dec 16 '18 at 16:23
$begingroup$
@user602338 You talk about completing the square in your question... are you sure it wasn't supposed to be squared instead of cubed?
$endgroup$
– Ovi
Dec 16 '18 at 16:23
$begingroup$
@Ovi Sorry but I really don't know how! Please help me with that!
$endgroup$
– user602338
Dec 16 '18 at 16:43
|
show 2 more comments
$begingroup$
I want to write y in terms of x. How do I do that?
$endgroup$
– user602338
Dec 16 '18 at 16:22
$begingroup$
@user602338 It's a way for solving a cubic equation. It's like the quadratic formula for cubic equations.
$endgroup$
– Ovi
Dec 16 '18 at 16:22
$begingroup$
@user602338 You can do that using Cardano's method brilliant.org/wiki/cardano-method
$endgroup$
– Ovi
Dec 16 '18 at 16:23
$begingroup$
@user602338 You talk about completing the square in your question... are you sure it wasn't supposed to be squared instead of cubed?
$endgroup$
– Ovi
Dec 16 '18 at 16:23
$begingroup$
@Ovi Sorry but I really don't know how! Please help me with that!
$endgroup$
– user602338
Dec 16 '18 at 16:43
$begingroup$
I want to write y in terms of x. How do I do that?
$endgroup$
– user602338
Dec 16 '18 at 16:22
$begingroup$
I want to write y in terms of x. How do I do that?
$endgroup$
– user602338
Dec 16 '18 at 16:22
$begingroup$
@user602338 It's a way for solving a cubic equation. It's like the quadratic formula for cubic equations.
$endgroup$
– Ovi
Dec 16 '18 at 16:22
$begingroup$
@user602338 It's a way for solving a cubic equation. It's like the quadratic formula for cubic equations.
$endgroup$
– Ovi
Dec 16 '18 at 16:22
$begingroup$
@user602338 You can do that using Cardano's method brilliant.org/wiki/cardano-method
$endgroup$
– Ovi
Dec 16 '18 at 16:23
$begingroup$
@user602338 You can do that using Cardano's method brilliant.org/wiki/cardano-method
$endgroup$
– Ovi
Dec 16 '18 at 16:23
$begingroup$
@user602338 You talk about completing the square in your question... are you sure it wasn't supposed to be squared instead of cubed?
$endgroup$
– Ovi
Dec 16 '18 at 16:23
$begingroup$
@user602338 You talk about completing the square in your question... are you sure it wasn't supposed to be squared instead of cubed?
$endgroup$
– Ovi
Dec 16 '18 at 16:23
$begingroup$
@Ovi Sorry but I really don't know how! Please help me with that!
$endgroup$
– user602338
Dec 16 '18 at 16:43
$begingroup$
@Ovi Sorry but I really don't know how! Please help me with that!
$endgroup$
– user602338
Dec 16 '18 at 16:43
|
show 2 more comments
$begingroup$
I don't understand $x_1=x_2$ here. What do you want to ask? If $2=2$ then what?
$endgroup$
– Dietrich Burde
Dec 16 '18 at 16:15
$begingroup$
Oh i m blind! Sorry .edited thnks
$endgroup$
– user602338
Dec 16 '18 at 16:16
$begingroup$
Your question doesn't make much sense. What do you want to know, if you can express $y$ in terms of $x$ as a function ?
$endgroup$
– Rebellos
Dec 16 '18 at 16:17
$begingroup$
Further to @DietrichBurde's question, you have $+x$ in the question but $+2x$ in the body.
$endgroup$
– J.G.
Dec 16 '18 at 16:17
$begingroup$
Check if we have 2 equal x , then y woudl be equal for them
$endgroup$
– user602338
Dec 16 '18 at 16:18