MATLAB: Can MATLAB solve 3 order equation including square root of the variable

MATLABsolve equation

Hi there,
I am using MATLAB to solve the below equation to obtain x. All the other variables are known.
The code used is shown below.
if true
  % code
syms a b c d e f g R x
eqn = a*x^3 + b*sqrt(R^2-x^2)*x^2 + c*x^2 + d*sqrt(R^2-x^2)*x...
  + e*x + f*sqrt(R^2-x^2) + g == 0;
sol = solve(eqn, x, 'ReturnConditions', true)
end
Then I have the results:
>> sol.conditions
=
(d^2 – 4*b*f)^(1/2) ~= d + 2*b*z1 & d + 2*b*z1 + (d^2 – 4*b*f)^(1/2) ~= 0 & a^2 + b^2 ~= 0 & R^2*b^2*z1^4 + 2*R^2*b*d*z1^3 + 2*R^2*b*f*z1^2 + R^2*d^2*z1^2 + 2*R^2*d*f*z1 + R^2*f^2 == a^2*z1^6 + 2*a*c*z1^5 + 2*a*e*z1^4 + 2*a*g*z1^3 + b^2*z1^6 + 2*b*d*z1^5 + 2*b*f*z1^4 + c^2*z1^4 + 2*c*e*z1^3 + 2*c*g*z1^2 + d^2*z1^4 + 2*d*f*z1^3 + e^2*z1^2 + 2*e*g*z1 + f^2*z1^2 + g^2 & (signIm(((a*z1^3 + c*z1^2 + e*z1 + g)*1i)/(b*z1^2 + d*z1 + f)) == -1 | a*z1^3 + c*z1^2 + e*z1 + g == 0) & b ~= 0
>> sol.parameters
=
z1
>> sol.x
ans =
z1
Does this mean there is no solution for the equation, please?
Thanks.

Best Answer

Hello Huan,
I don't believe a symbolic solution is possible but if you can live with a numerical solution, then you can take all the terms involving sqrt(R^2-x^2) over to the right hand side, square both sides, bring the right hand side back and obtain a sixth degree polynomial
(a*x^3 + c*x^2 + e*x + g)^2 - (b*x^2 + d*x + f)^2*(R^2-x^2) = 0.
One can square this out and find the coefficients by hand but it helps to have the symbolic results
syms a b c d e f g R x
z = (a*x^3 + c*x^2 + e*x + g)^2 - (b*x^2 + d*x + f)^2*(R^2-x^2)
z = collect(expand(z))
which gives
z = (a^2 + b^2)*x^6 + (2*a*c + 2*b*d)*x^5  etc. etc. (results below)
The coefficients are denoted by q below and the solutions are the roots of that polynomial. Leaving syms at this point and putting in arbitrary values for the constants, then in regular variables the solution is
a = 1;  b = 2;  c = 3;  d = 4;  e = 5;  f = 6;  g = 7;  R = 8;
q6 = a^2 + b^2;
q5 = 2*a*c + 2*b*d;
q4 = - R^2*b^2 + 2*f*b + c^2 + d^2 + 2*a*e;
q3 = - 2*b*d*R^2 + 2*a*g + 2*c*e + 2*d*f;
q2 = - R^2*d^2 - 2*b*R^2*f + e^2 + f^2 + 2*c*g;
q1 = - 2*d*f*R^2 + 2*e*g;
q0 = - R^2*f^2 + g^2;
x = roots([q6 q5 q4 q3 q2 q1 q0])
x =
 6.9305 + 0.0000i
-7.3609 + 0.0000i
-0.9906 + 1.4981i
-0.9906 - 1.4981i
-0.9942 + 1.3238i
-0.9942 - 1.3238i
In this case two solutions are real, four are complex. Since the original equation was squared, the solutions can involve either positive sqrt(R^2-x^2) (three solutions) or negative sqrt(R^2-x^2) (three solutions). If you check the original equation
(a*x.^3 + c*x.^2 + e*x + g) - (b*x.^2 + d*x + f).*sqrt(R^2-x.^2)    % should be 0 for pos. sqrt
(a*x.^3 + c*x.^2 + e*x + g) + (b*x.^2 + d*x + f).*sqrt(R^2-x.^2)    % should be 0 for neg. sqrt
the results show that in this example solutions 1,5,6 use positive sqrt and solutions 2,3,4 use negative sqrt. If you want a solution with positive square root and x real, then the first root x = 6.9305 gets the job done. In general there is no guarantee that there is a solution with x real.
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