Help solving complicated polynomial equation
up vote
2
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I am trying to solve the following two equations, but I am having troubles doing so.
Equation 1 :
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3) ;
Solve[eq1[n] == 0, n]
If I can find an "n", I would then plug it into the following equation and solve for V.
eq2[n_] := 1024/5*n + 7133.17 n^3 -
0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[nFound] == 0, V]
equation-solving
asked 2 hours ago
james
758519
add a comment |
up vote
2
down vote
favorite
I am trying to solve the following two equations, but I am having troubles doing so.
Equation 1 :
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3) ;
Solve[eq1[n] == 0, n]
If I can find an "n", I would then plug it into the following equation and solve for V.
eq2[n_] := 1024/5*n + 7133.17 n^3 -
0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[nFound] == 0, V]
equation-solving
asked 2 hours ago
james
758519
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am trying to solve the following two equations, but I am having troubles doing so.
Equation 1 :
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3) ;
Solve[eq1[n] == 0, n]
If I can find an "n", I would then plug it into the following equation and solve for V.
eq2[n_] := 1024/5*n + 7133.17 n^3 -
0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[nFound] == 0, V]
equation-solving
asked 2 hours ago
james
758519
I am trying to solve the following two equations, but I am having troubles doing so.
Equation 1 :
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3) ;
Solve[eq1[n] == 0, n]
If I can find an "n", I would then plug it into the following equation and solve for V.
eq2[n_] := 1024/5*n + 7133.17 n^3 -
0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[nFound] == 0, V]
equation-solving
equation-solving
asked 2 hours ago
james
758519
asked 2 hours ago
james
758519
asked 2 hours ago
james
758519
asked 2 hours ago
james
758519
asked 2 hours ago
james
758519
758519
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Sorry, I am not very sure what you want to ask. Maybe you want to substitute the solution of the first equation into the second equation? Then you can do this.
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3);
ansList = Values@Solve[eq1[n] == 0, n]
eq2[n_] :=
1024/5*n + 7133.17 n^3 - 0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[#] == 0, V] & /@ ansList
You also can use the following code to get the value of n.
sol=Solve[eq1[n] == 0, n]
nList=n/.sol
That's all. Please enjoy the fun of Mathematica.
answered 1 hour ago
J.W Kang
635
Thanks ! This was exactly what I was looking for.
– james
1 hour ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Sorry, I am not very sure what you want to ask. Maybe you want to substitute the solution of the first equation into the second equation? Then you can do this.
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3);
ansList = Values@Solve[eq1[n] == 0, n]
eq2[n_] :=
1024/5*n + 7133.17 n^3 - 0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[#] == 0, V] & /@ ansList
You also can use the following code to get the value of n.
sol=Solve[eq1[n] == 0, n]
nList=n/.sol
That's all. Please enjoy the fun of Mathematica.
answered 1 hour ago
J.W Kang
635
Thanks ! This was exactly what I was looking for.
– james
1 hour ago
add a comment |
up vote
2
down vote
accepted
Sorry, I am not very sure what you want to ask. Maybe you want to substitute the solution of the first equation into the second equation? Then you can do this.
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3);
ansList = Values@Solve[eq1[n] == 0, n]
eq2[n_] :=
1024/5*n + 7133.17 n^3 - 0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[#] == 0, V] & /@ ansList
You also can use the following code to get the value of n.
sol=Solve[eq1[n] == 0, n]
nList=n/.sol
That's all. Please enjoy the fun of Mathematica.
answered 1 hour ago
J.W Kang
635
Thanks ! This was exactly what I was looking for.
– james
1 hour ago
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Sorry, I am not very sure what you want to ask. Maybe you want to substitute the solution of the first equation into the second equation? Then you can do this.
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3);
ansList = Values@Solve[eq1[n] == 0, n]
eq2[n_] :=
1024/5*n + 7133.17 n^3 - 0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[#] == 0, V] & /@ ansList
You also can use the following code to get the value of n.
sol=Solve[eq1[n] == 0, n]
nList=n/.sol
That's all. Please enjoy the fun of Mathematica.
answered 1 hour ago
J.W Kang
635
Sorry, I am not very sure what you want to ask. Maybe you want to substitute the solution of the first equation into the second equation? Then you can do this.
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3);
ansList = Values@Solve[eq1[n] == 0, n]
eq2[n_] :=
1024/5*n + 7133.17 n^3 - 0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[#] == 0, V] & /@ ansList
You also can use the following code to get the value of n.
sol=Solve[eq1[n] == 0, n]
nList=n/.sol
That's all. Please enjoy the fun of Mathematica.
answered 1 hour ago
J.W Kang
635
answered 1 hour ago
J.W Kang
635
answered 1 hour ago
J.W Kang
635
answered 1 hour ago
J.W Kang
635
635
Thanks ! This was exactly what I was looking for.
– james
1 hour ago
add a comment |
Thanks ! This was exactly what I was looking for.
– james
1 hour ago
Thanks ! This was exactly what I was looking for.
– james
1 hour ago
Thanks ! This was exactly what I was looking for.
– james
1 hour ago
add a comment |
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