Solution to set of cubic equations
$begingroup$
Let us consider the following set of equations:
$$a|x_i|^6 + b|x_i|^4 + c|x_i|^2 + d = 0,$$
where $i=1,2,3$ and $a,b,c,d$ are real, non-zero numbers. We can easily obtain three different solutions for the complex $x_i$, i.e., $x_1neq x_2neq x_3$.
Let us now add another term to the above equations, which depends non-trivially on $x_j$ and $x_k$:
$$a|x_i|^6 + b|x_i|^4 + c|x_i|^2 + d + ex_jx_k = 0,$$
where $i,j,k=1,2,3$ and $ineq jneq k$ and $e$ is a real, non-zero number.
Are there still three different solutions obtainable, i.e., $x_1neq x_2neq x_3$? If so, how do we obtain these solutions?
polynomials cubic-equations
asked Dec 7 '18 at 22:51
LCFLCF
1327
$endgroup$
add a comment |
$begingroup$
Let us consider the following set of equations:
$$a|x_i|^6 + b|x_i|^4 + c|x_i|^2 + d = 0,$$
where $i=1,2,3$ and $a,b,c,d$ are real, non-zero numbers. We can easily obtain three different solutions for the complex $x_i$, i.e., $x_1neq x_2neq x_3$.
Let us now add another term to the above equations, which depends non-trivially on $x_j$ and $x_k$:
$$a|x_i|^6 + b|x_i|^4 + c|x_i|^2 + d + ex_jx_k = 0,$$
where $i,j,k=1,2,3$ and $ineq jneq k$ and $e$ is a real, non-zero number.
Are there still three different solutions obtainable, i.e., $x_1neq x_2neq x_3$? If so, how do we obtain these solutions?
polynomials cubic-equations
asked Dec 7 '18 at 22:51
LCFLCF
1327
$endgroup$
add a comment |
$begingroup$
Let us consider the following set of equations:
$$a|x_i|^6 + b|x_i|^4 + c|x_i|^2 + d = 0,$$
where $i=1,2,3$ and $a,b,c,d$ are real, non-zero numbers. We can easily obtain three different solutions for the complex $x_i$, i.e., $x_1neq x_2neq x_3$.
Let us now add another term to the above equations, which depends non-trivially on $x_j$ and $x_k$:
$$a|x_i|^6 + b|x_i|^4 + c|x_i|^2 + d + ex_jx_k = 0,$$
where $i,j,k=1,2,3$ and $ineq jneq k$ and $e$ is a real, non-zero number.
Are there still three different solutions obtainable, i.e., $x_1neq x_2neq x_3$? If so, how do we obtain these solutions?
polynomials cubic-equations
asked Dec 7 '18 at 22:51
LCFLCF
1327
$endgroup$
Let us consider the following set of equations:
$$a|x_i|^6 + b|x_i|^4 + c|x_i|^2 + d = 0,$$
where $i=1,2,3$ and $a,b,c,d$ are real, non-zero numbers. We can easily obtain three different solutions for the complex $x_i$, i.e., $x_1neq x_2neq x_3$.
Let us now add another term to the above equations, which depends non-trivially on $x_j$ and $x_k$:
$$a|x_i|^6 + b|x_i|^4 + c|x_i|^2 + d + ex_jx_k = 0,$$
where $i,j,k=1,2,3$ and $ineq jneq k$ and $e$ is a real, non-zero number.
Are there still three different solutions obtainable, i.e., $x_1neq x_2neq x_3$? If so, how do we obtain these solutions?
polynomials cubic-equations
polynomials cubic-equations
asked Dec 7 '18 at 22:51
LCFLCF
1327
asked Dec 7 '18 at 22:51
LCFLCF
1327
asked Dec 7 '18 at 22:51
LCFLCF
1327
asked Dec 7 '18 at 22:51
LCFLCF
1327
asked Dec 7 '18 at 22:51
LCFLCF
1327
1327
add a comment |
add a comment |
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