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.
