how to solve equations ?
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hello everybody,
I have two equations with (d1,d2,L,x) variables and I want to find x in terms of (d1,d2,L) variables .I used "solve" function to do this but it gave me x in term of (d1,d2,L,z) where I have not used "z" variable in my equations, and I can't understand what is this.
This is what I did:
syms x d1 d2 l;
y1=((d2 + x)^2 - 2*l^2 + (d1 - x)^2)/(2*l*(d1 + x)^2*(d2 + x)) + ((d1^2 + 2*x*d1 + l^2)^2*(d2^2 + 2*x*d2 + l^2)^2)/(16*l^4*(d1 + x)^2*(d2 + x)^2);%y1=0
y2=((d1^2 + 2*x*d1 + l^2)^2/(4*l^2*(d1 + x)^2) - 1)*((d2^2 + 2*x*d2 + l^2)^2/(4*l^2*(d2 + x)^2) - 1); %y2=0
final_equation=y1-y2;
x=solve(final_equation,x)
any help in this regard will be appreciated
2 Comments
Walter Roberson
on 25 Dec 2012
Please read the guide to tags and retag this question. see http://www.mathworks.co.uk/matlabcentral/answers/43073-a-guide-to-tags
AYAH
on 25 Dec 2012
Answers (1)
Walter Roberson
on 25 Dec 2012
0 votes
The solution for that is a quartic. There are analytic solutions, but they are very long to write out. MATLAB shows the short form of them by returning a RootOf() placeholder. RootOf(f(z),z) means "the values of z such that f(z) returns 0".
You can tell solve() to give you the complete analytic solution using the maximum degree option, but beware that it runs to numerous pages and is basically incomprehensible.
2 Comments
AYAH
on 26 Dec 2012
Edited: Walter Roberson
on 26 Dec 2012
Walter Roberson
on 26 Dec 2012
Odd. Which MATLAB version are you using?
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