how to solve equations ?

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AYAH 2012-12-25

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https://ww2.mathworks.cn/matlabcentral/answers/57379-how-to-solve-equations

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

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Walter Roberson 2012-12-25

Please read the guide to tags and retag this question. see http://www.mathworks.co.uk/matlabcentral/answers/43073-a-guide-to-tags

AYAH 2012-12-25

yes it appear within a RootOf()

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Answer 1

Walter Roberson 2012-12-25

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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.

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AYAH 2012-12-26

编辑：Walter Roberson 2012-12-26

在 MATLAB Online 中打开

I have tried maximum degree option but I got an error:

??? Error using ==> solve>getEqns at 202
' 4 ' is not a valid expression or equation.

this was my code:

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;
S=solve(final_equation,x,'MaxDegree',4);
pretty(S)