Rid polynomial of square-root



Maxima has at least two alternative solve functions. The %solve function defined in the optional
to_poly_solver package can solve your equation.

 (%i1) load("to_poly_solver")$

 (%i2) sol : %solve(x = sqrt(x^2+1) - a,x);
 (%o2) %union(%if((-%pi/2<parg((a^2+1)/a)) %and (parg((a^2+1)/a)<=%pi/2),[x=-(a^2-1)/(2*a)],%union()))

parg is the complex number argument that is in (-pi, pi]. We can substitute values for a to get 
specific solutions:

 (%i3) subst(a=42,sol);
 (%o3) %union([x=-1763/84])

When a = -1, the solution set is empty:

 (%i4) subst(a=-1,sol);
 (%o4) %union()

When a = -1, squaring the equation makes the solution set larger. The to_poly_solver tries to expunge
spurious solutions (it's a bug if it doesn't).

--Barton

Author of to_poly_solver
http://www.unk.edu/uploadedFiles/facstaff/profiles/willisb/solve-talk(3).pdf

-----maxima-bounces at math.utexas.edu wrote: -----


>You?can?solve?it?"by?hand"?using?maxima.??Where?does?where?b?come?into?it?
>
>(%i1)?x=sqrt(x^2+1)-a;
>???????????????????????????????????????2
>(%o1)????????????????????????x?=?sqrt(x??+?1)?-?a
>(%i2)?%+a;
>???????????????????????????????????????????2
>(%o2)????????????????????????x?+?a?=?sqrt(x??+?1)
>(%i3)?lhs(%)^2=rhs(%)^2;
>??????????????????????????????????????2????2
>(%o3)??????????????????????????(x?+?a)??=?x??+?1
>(%i4)?expand(%);
>????????????????????????????2????????????2????2
>(%o4)??????????????????????x??+?2?a?x?+?a??=?x??+?1
>(%i5)?lhs(%)-rhs(%)=0;
>???????????????????????????????????????2
>(%o5)?????????????????????????2?a?x?+?a??-?1?=?0
>(%i6)?solve(%,x);
>????????????????????????????????????????2
>???????????????????????????????????????a??-?1
>(%o6)???????????????????????????[x?=?-?------]
>????????????????????????????????????????2?a
>
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