Solve quadratic equations without individually specifying cofficients

Question

I am trying to solve equations like:

3*(3x-12)/(x+3)-2*(2x+3)/(3x-1) = 5

This is the code that I use:

eqn1 = 3*(3*X-12)/(X+3)-2*(2*X+3)/(3*X-1) == 5;
sol = solve(eqn1, X);
xSol = sol.X

This is the error that I get:

Error using sym/subsref
Too many output arguments.

Please explain what changes between your equations (give some examples of different possibilities). How do you get the inputs? As a vector of coefficients? As a symbolic string? Do you expect to get all possible solutions or just one? — Dev-iL, Jul 05 '18 at 07:56

score 3 · Accepted Answer · answered Jul 05 '18 at 07:48

The first thing I'd suggest is performing a graphical solution:

% Define the function:
f = @(X)3*(3*X-12)./(X+3)-2*(2*X+3)./(3*X-1)-5;
% Plot the function (solve graphically):
x = -30:0.1:30;
figure(); plot(x,f(x)); grid on; grid minor;

This function has vertical asymptotes and a horizontal asymptote, at x=-3, x=1/3 and y=8/3 (finding this is left as an exercise to the reader). Let's add them to the chart and zoom to the y-vicinity of 0:

hold on; plot([-3, -3, NaN, 1/3, 1/3], 600*[-1, 1, NaN, -1, 1],'--r');
plot([-30, 30], 8/3*[1 1], '--m'); ylim([-10 10]);

There appear to be two solution, one between the vertical asymptotes and another to the right of the right asymptote. We can define these regions for fzero:

% Find zeros:
z = [ fzero(f, [-3+eps(3) 1/3-eps(1/3)] ),... First solution
      fzero(f, [1/3+eps, 30])];             % Second solution

(where 30 is some sufficiently large number) and we get:

z =

    0.1902   21.6848

Solve quadratic equations without individually specifying cofficients

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