roots of a several variable for the equation $y-3x^2-y^3=0$
$begingroup$
How can I get the roots of the next equation?
$$y-3x^2-y^3=0$$
I just dont get the same answer than my teacher:
$$x = frac {- sqrt2}{3(3^{1/4})}, y = frac{-2}{sqrt3}$$
several-complex-variables
edited Jan 8 at 18:30
gt6989b
35.2k22557
asked Jan 8 at 18:18
MoiraMoira
1
$endgroup$
add a comment |
$begingroup$
How can I get the roots of the next equation?
$$y-3x^2-y^3=0$$
I just dont get the same answer than my teacher:
$$x = frac {- sqrt2}{3(3^{1/4})}, y = frac{-2}{sqrt3}$$
several-complex-variables
edited Jan 8 at 18:30
gt6989b
35.2k22557
asked Jan 8 at 18:18
MoiraMoira
1
$endgroup$
2
$begingroup$
That equation has many roots
$endgroup$
– EuxhenH
Jan 8 at 18:22
4
$begingroup$
There had to be something else in the question. Maybe it was a system of two equations, not just this one equation by itself?
$endgroup$
– zipirovich
Jan 8 at 18:22
$begingroup$
$x=0$, $y=0$, is just as valid.
$endgroup$
– Peter Foreman
Jan 8 at 18:27
add a comment |
$begingroup$
How can I get the roots of the next equation?
$$y-3x^2-y^3=0$$
I just dont get the same answer than my teacher:
$$x = frac {- sqrt2}{3(3^{1/4})}, y = frac{-2}{sqrt3}$$
several-complex-variables
edited Jan 8 at 18:30
gt6989b
35.2k22557
asked Jan 8 at 18:18
MoiraMoira
1
$endgroup$
How can I get the roots of the next equation?
$$y-3x^2-y^3=0$$
I just dont get the same answer than my teacher:
$$x = frac {- sqrt2}{3(3^{1/4})}, y = frac{-2}{sqrt3}$$
several-complex-variables
several-complex-variables
edited Jan 8 at 18:30
gt6989b
35.2k22557
asked Jan 8 at 18:18
MoiraMoira
1
edited Jan 8 at 18:30
gt6989b
35.2k22557
asked Jan 8 at 18:18
MoiraMoira
1
edited Jan 8 at 18:30
gt6989b
35.2k22557
edited Jan 8 at 18:30
gt6989b
35.2k22557
edited Jan 8 at 18:30
gt6989b
35.2k22557
35.2k22557
asked Jan 8 at 18:18
MoiraMoira
1
asked Jan 8 at 18:18
MoiraMoira
1
asked Jan 8 at 18:18
MoiraMoira
1
1
2
$begingroup$
That equation has many roots
$endgroup$
– EuxhenH
Jan 8 at 18:22
4
$begingroup$
There had to be something else in the question. Maybe it was a system of two equations, not just this one equation by itself?
$endgroup$
– zipirovich
Jan 8 at 18:22
$begingroup$
$x=0$, $y=0$, is just as valid.
$endgroup$
– Peter Foreman
Jan 8 at 18:27
add a comment |
2
$begingroup$
That equation has many roots
$endgroup$
– EuxhenH
Jan 8 at 18:22
4
$begingroup$
There had to be something else in the question. Maybe it was a system of two equations, not just this one equation by itself?
$endgroup$
– zipirovich
Jan 8 at 18:22
$begingroup$
$x=0$, $y=0$, is just as valid.
$endgroup$
– Peter Foreman
Jan 8 at 18:27
2
2
$begingroup$
That equation has many roots
$endgroup$
– EuxhenH
Jan 8 at 18:22
$begingroup$
That equation has many roots
$endgroup$
– EuxhenH
Jan 8 at 18:22
4
4
$begingroup$
There had to be something else in the question. Maybe it was a system of two equations, not just this one equation by itself?
$endgroup$
– zipirovich
Jan 8 at 18:22
$begingroup$
There had to be something else in the question. Maybe it was a system of two equations, not just this one equation by itself?
$endgroup$
– zipirovich
Jan 8 at 18:22
$begingroup$
$x=0$, $y=0$, is just as valid.
$endgroup$
– Peter Foreman
Jan 8 at 18:27
$begingroup$
$x=0$, $y=0$, is just as valid.
$endgroup$
– Peter Foreman
Jan 8 at 18:27
add a comment |
1 Answer
1
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$begingroup$
Any point on the graph will satisfy the equation. You might wish to verify that your teacher's solution lies on the graph by squaring the $x$, cubing the $y$ and taking the designated combination.
answered Jan 8 at 18:30
bouncebackbounceback
444212
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Any point on the graph will satisfy the equation. You might wish to verify that your teacher's solution lies on the graph by squaring the $x$, cubing the $y$ and taking the designated combination.
answered Jan 8 at 18:30
bouncebackbounceback
444212
$endgroup$
add a comment |
$begingroup$
Any point on the graph will satisfy the equation. You might wish to verify that your teacher's solution lies on the graph by squaring the $x$, cubing the $y$ and taking the designated combination.
answered Jan 8 at 18:30
bouncebackbounceback
444212
$endgroup$
add a comment |
$begingroup$
Any point on the graph will satisfy the equation. You might wish to verify that your teacher's solution lies on the graph by squaring the $x$, cubing the $y$ and taking the designated combination.
answered Jan 8 at 18:30
bouncebackbounceback
444212
$endgroup$
Any point on the graph will satisfy the equation. You might wish to verify that your teacher's solution lies on the graph by squaring the $x$, cubing the $y$ and taking the designated combination.
answered Jan 8 at 18:30
bouncebackbounceback
444212
answered Jan 8 at 18:30
bouncebackbounceback
444212
answered Jan 8 at 18:30
bouncebackbounceback
444212
answered Jan 8 at 18:30
bouncebackbounceback
444212
444212
add a comment |
add a comment |
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$begingroup$
That equation has many roots
$endgroup$
– EuxhenH
Jan 8 at 18:22
4
$begingroup$
There had to be something else in the question. Maybe it was a system of two equations, not just this one equation by itself?
$endgroup$
– zipirovich
Jan 8 at 18:22
$begingroup$
$x=0$, $y=0$, is just as valid.
$endgroup$
– Peter Foreman
Jan 8 at 18:27