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Solve me numerically 0=3+x^2+5*x^3 using jlFriCAS
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Numerical solution for the equation using jlFriCAS yields the following roots:
These values were obtained by calling complexSolve(5*x^3 + x^2 + 3 = 0, 1.e-10) within the FriCAS environment.
Can you return results but not in LaTeX form
The numerical solutions for in plain text are:
Real Root: x1 = -0.9156445838
Complex Roots: x2 = 0.3578222919 - 0.7261124531i x3 = 0.3578222919 + 0.7261124531i
What is the documentation of complexSolve please
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*Viewed [numsolve.spad](file:///home/greg/Git/jlfricas/src/algebra/numsolve.spad) *
The documentation for complexSolve (from the FloatingComplexPackage) describes it as a function for finding approximate complex solutions to systems of polynomial equations or rational functions.
complexSolve(p, eps): Finds all complex solutions of the rational function p (or polynomial) with respect to its variable, using precision eps.complexSolve(eq, eps): Finds all complex solutions of the equation eq.complexSolve(lp, eps): Finds all complex solutions for a list of rational functions lp.complexSolve(leq, eps): Finds all complex solutions for a list of equations leq.eps):1.e-10), the results are returned as Complex Floats.1/10^10), the results are returned as Complex Rational Numbers."This is a package for the approximation of complex solutions for systems of equations of rational functions with complex rational coefficients. The results are expressed as either complex rational numbers or complex floats depending on the type of the precision parameter..."
The function internally uses the ComplexRootPackage for univariate cases and a Groebner-basis-based solver (GroebnerSolve) for multivariate systems to isolate roots before refining them to the requested precision.
Solve me 0=3+x^2+5*x^3 using jlFriCAS
Running MCP tool
The numerical solutions for the equation are:
Real Root: x = -0.9156445838
Complex Roots: x = 0.3578222919 - 0.7261124531i x = 0.3578222919 + 0.7261124531i
These results were calculated with a precision of 10^-10.
Can you solve it symbolically?
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The symbolic solutions for the equation 5x^3 + x^2 + 3 = 0 involve nested radicals. Here they are as provided by jlFriCAS:
Root 1:
+-----------------+2
| +----+
+---+ |45 \|2029 - 2027
(- \|- 3+1)3|-----------------
\| 2
+
+-----------------+
| +----+
+---+ |45 \|2029 - 2027
(- \|- 3-1)3|----------------- - 2
\| 2
/
+-----------------+
| +----+
+---+ |45 \|2029 - 2027
(15 \|- 3+15)3|-----------------
\| 2
Root 2:
+-----------------+2
| +----+
+---+ |45 \|2029 - 2027
(- \|- 3-1)3|-----------------
\| 2
+
+-----------------+
| +----+
+---+ |45 \|2029 - 2027
(- \|- 3+1)3|----------------- + 2
\| 2
/
+-----------------+
| +----+
+---+ |45 \|2029 - 2027
(15 \|- 3-15)3|-----------------
\| 2
Root 3:
+-----------------+2 +-----------------+
| +----+ | +----+
|45 \|2029 - 2027 |45 \|2029 - 2027
3|----------------- - 3|----------------- + 1
\| 2 \| 2
x = ------------------------------------------------
+-----------------+
| +----+
|45 \|2029 - 2027
15 3|-----------------
\| 2
These solutions are derived using the cubic formula. The expression \|- 3 refers to the square root of -3 (proportional to the imaginary unit i).