Solver added a variable?

 syms x y z;
solve(z==14.78+6.26*x+(-2.175)*y+(-0.9272)*x^2+(-1.357)*x*y+1.316*y^2+1.504*x^3+0.08749*x^2*y+0.3966*x*y^2+(-0.4403)*y^3,x)

This give me a answer of:

ans =

 root(z1^3 + (125*z1^2*((8749*y)/100000 - 1159/1250))/188 + (125*z1*(- (1357*y)/1000 + (1983*y^2)/5000 + 313/50))/188 - (125*z)/188 - (2175*y)/1504 - (4403*y^3)/15040 + (7*y^2)/8 + 3695/376, z1, 1)
 root(z1^3 + (125*z1^2*((8749*y)/100000 - 1159/1250))/188 + (125*z1*(- (1357*y)/1000 + (1983*y^2)/5000 + 313/50))/188 - (125*z)/188 - (2175*y)/1504 - (4403*y^3)/15040 + (7*y^2)/8 + 3695/376, z1, 2)
 root(z1^3 + (125*z1^2*((8749*y)/100000 - 1159/1250))/188 + (125*z1*(- (1357*y)/1000 + (1983*y^2)/5000 + 313/50))/188 - (125*z)/188 - (2175*y)/1504 - (4403*y^3)/15040 + (7*y^2)/8 + 3695/376, z1, 3)

What is the 'z1' in the answer? Where did it come from, and what do I do with it?

Thanks in advance for the help!

Answer 1

[Walter Roberson]

The general form is

root(expression_in_variable, variable, number)

root(expression_in_variable, variable) is a place holder for the complete set of values of the named variable such that the expression in the variable becomes 0 -- that is, root() stands in for the roots of the given expression.

When a number is specified, then one particular root is to be selected. You are dealing with a cubic polynomial, so there are three roots (that are possibly complex) and the selectors 1, 2, and 3 designate which of the roots are being referred to.

The numbering of the roots is not documented. For any given expression, the numbering is consistent, but it might be difficult to work out the rules about how it decides which of the roots is associated with which index.

MATLAB introduces a new variable for this purpose because in general the variable you were trying to solve() for is an expression on one of these root() values, rather than being one of the root values directly.