# Maxima Help

## Numerical Solutions

### find_root(function, variable, lower bound, upper bound);

find_root(x^3-3*x+1,x,0,1);

will give you 0.34729635533386

The find_root() command requires a left bound and a right bound and an x-intercept in-between them. If there are multiple x-intercepts between the left and right bounds, then the results are unpredictable.

## Algebraic Solutions

### solve(equations, variables);

The variable may be omitted if it is obvious what you want. It says equations, but you can leave off the = 0.

solve(x^3-3*x+1);

will give you

- \( \displaystyle x={\left( \frac{\sqrt{3}\,i}{2}-\frac{1}{2}\right) }^{\frac{2}{3}}+\left( -\frac{\sqrt{3}\,i}{2}-\frac{1}{2}\right) \,{\left( \frac{\sqrt{3}\,i}{2}-\frac{1}{2}\right) }^{\frac{1}{3}}\)
- \( \displaystyle x={\left( \frac{\sqrt{3}\,i}{2}-\frac{1}{2}\right) }^{\frac{4}{3}}+\frac{-\frac{\sqrt{3}\,i}{2}-\frac{1}{2}}{{\left( \frac{\sqrt{3}\,i}{2}-\frac{1}{2}\right) }^{\frac{1}{3}}}\)
- \( \displaystyle x={\left( \frac{\sqrt{3}\,i}{2}-\frac{1}{2}\right) }^{\frac{1}{3}}+\frac{1}{{\left( \frac{\sqrt{3}\,i}{2}-\frac{1}{2}\right) }^{\frac{1}{3}}}\)

None of the results from solve are useful, but solve() is usually better when it works as it will give you all of the solutions it can find (except for ones involving periodic functions like the trigonometric functions).

You can also solve a system of equations with the solve() command, but that's not useful for this project.