How to avoid introducing a spurious solution?

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Suppose I have



$sqrtr-x-fracxsqrtr-x-o=0$
with $r,x$ and $o$ real.



To solve I multiply by $sqrtr-x$.
I assume therefore $r>x$



$r-x-x-osqrtr-x=0$



$r-2x=osqrtr-x$



Squaring



$(r-2x)^2=o^2(r-x)$



This is a second order equation,whose solutions are



$x= fracr2 -fraco^28 pm fracosqrto^2+8r8$



However, for instance, for $r=1$ and $o=1/2$ I get two solutions. For both $x<1$, however only one is actually a solution.



How can I avoid to introduce a spurious solution?




quadratics




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  • I edited the question,now it should be clearer
    – Fabio Dalla Libera
    Aug 23 at 7:34






  • 1




    The solutions you found solve $r-2x=osqrtr-x$ and also $r-2x=-osqrtr-x$.
    – Did
    Aug 23 at 7:35










  • @Did Consider posting as answer...
    – user202729
    Aug 23 at 7:35














up vote
0
down vote

favorite












Suppose I have



$sqrtr-x-fracxsqrtr-x-o=0$
with $r,x$ and $o$ real.



To solve I multiply by $sqrtr-x$.
I assume therefore $r>x$



$r-x-x-osqrtr-x=0$



$r-2x=osqrtr-x$



Squaring



$(r-2x)^2=o^2(r-x)$



This is a second order equation,whose solutions are



$x= fracr2 -fraco^28 pm fracosqrto^2+8r8$



However, for instance, for $r=1$ and $o=1/2$ I get two solutions. For both $x<1$, however only one is actually a solution.



How can I avoid to introduce a spurious solution?




quadratics




share|cite|improve this question






















  • I edited the question,now it should be clearer
    – Fabio Dalla Libera
    Aug 23 at 7:34






  • 1




    The solutions you found solve $r-2x=osqrtr-x$ and also $r-2x=-osqrtr-x$.
    – Did
    Aug 23 at 7:35










  • @Did Consider posting as answer...
    – user202729
    Aug 23 at 7:35












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Suppose I have



$sqrtr-x-fracxsqrtr-x-o=0$
with $r,x$ and $o$ real.



To solve I multiply by $sqrtr-x$.
I assume therefore $r>x$



$r-x-x-osqrtr-x=0$



$r-2x=osqrtr-x$



Squaring



$(r-2x)^2=o^2(r-x)$



This is a second order equation,whose solutions are



$x= fracr2 -fraco^28 pm fracosqrto^2+8r8$



However, for instance, for $r=1$ and $o=1/2$ I get two solutions. For both $x<1$, however only one is actually a solution.



How can I avoid to introduce a spurious solution?




quadratics




share|cite|improve this question














Suppose I have



$sqrtr-x-fracxsqrtr-x-o=0$
with $r,x$ and $o$ real.



To solve I multiply by $sqrtr-x$.
I assume therefore $r>x$



$r-x-x-osqrtr-x=0$



$r-2x=osqrtr-x$



Squaring



$(r-2x)^2=o^2(r-x)$



This is a second order equation,whose solutions are



$x= fracr2 -fraco^28 pm fracosqrto^2+8r8$



However, for instance, for $r=1$ and $o=1/2$ I get two solutions. For both $x<1$, however only one is actually a solution.



How can I avoid to introduce a spurious solution?





quadratics





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 23 at 7:55

























asked Aug 23 at 7:20









Fabio Dalla Libera

1275




1275











  • I edited the question,now it should be clearer
    – Fabio Dalla Libera
    Aug 23 at 7:34






  • 1




    The solutions you found solve $r-2x=osqrtr-x$ and also $r-2x=-osqrtr-x$.
    – Did
    Aug 23 at 7:35










  • @Did Consider posting as answer...
    – user202729
    Aug 23 at 7:35
















  • I edited the question,now it should be clearer
    – Fabio Dalla Libera
    Aug 23 at 7:34






  • 1




    The solutions you found solve $r-2x=osqrtr-x$ and also $r-2x=-osqrtr-x$.
    – Did
    Aug 23 at 7:35










  • @Did Consider posting as answer...
    – user202729
    Aug 23 at 7:35















I edited the question,now it should be clearer
– Fabio Dalla Libera
Aug 23 at 7:34




I edited the question,now it should be clearer
– Fabio Dalla Libera
Aug 23 at 7:34




1




1




The solutions you found solve $r-2x=osqrtr-x$ and also $r-2x=-osqrtr-x$.
– Did
Aug 23 at 7:35




The solutions you found solve $r-2x=osqrtr-x$ and also $r-2x=-osqrtr-x$.
– Did
Aug 23 at 7:35












@Did Consider posting as answer...
– user202729
Aug 23 at 7:35




@Did Consider posting as answer...
– user202729
Aug 23 at 7:35










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










The spurious solution is introduced by the squaring. Before squaring, $r-2x$ is constrained to have the sign of $o$ (as the square root is a positive number). When squaring, you drop this condition.



Alternative resolution:



Let $z:=sqrtr-x>0$.



The equation becomes



$$z-fracr-z^2z-o=0$$



or



$$2z^2-oz-r=0.$$



The only positive solutions in $z$ is



$$z=fracocolorgreen+sqrto^2+8r4.$$



It exist iff $o^2+8rge0$, and $x=r-z^2$ is implicitly $le r$.






share|cite|improve this answer


















  • 1




    thank you, this is a very elegant way
    – Fabio Dalla Libera
    Aug 23 at 7:56










  • How bad does that $o$ look there. :-) Indeed, the OP could employ another suitable symbol, alphabet etc.. Nice way +1
    – mrs
    Aug 23 at 8:03










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










The spurious solution is introduced by the squaring. Before squaring, $r-2x$ is constrained to have the sign of $o$ (as the square root is a positive number). When squaring, you drop this condition.



Alternative resolution:



Let $z:=sqrtr-x>0$.



The equation becomes



$$z-fracr-z^2z-o=0$$



or



$$2z^2-oz-r=0.$$



The only positive solutions in $z$ is



$$z=fracocolorgreen+sqrto^2+8r4.$$



It exist iff $o^2+8rge0$, and $x=r-z^2$ is implicitly $le r$.






share|cite|improve this answer


















  • 1




    thank you, this is a very elegant way
    – Fabio Dalla Libera
    Aug 23 at 7:56










  • How bad does that $o$ look there. :-) Indeed, the OP could employ another suitable symbol, alphabet etc.. Nice way +1
    – mrs
    Aug 23 at 8:03














up vote
1
down vote



accepted










The spurious solution is introduced by the squaring. Before squaring, $r-2x$ is constrained to have the sign of $o$ (as the square root is a positive number). When squaring, you drop this condition.



Alternative resolution:



Let $z:=sqrtr-x>0$.



The equation becomes



$$z-fracr-z^2z-o=0$$



or



$$2z^2-oz-r=0.$$



The only positive solutions in $z$ is



$$z=fracocolorgreen+sqrto^2+8r4.$$



It exist iff $o^2+8rge0$, and $x=r-z^2$ is implicitly $le r$.






share|cite|improve this answer


















  • 1




    thank you, this is a very elegant way
    – Fabio Dalla Libera
    Aug 23 at 7:56










  • How bad does that $o$ look there. :-) Indeed, the OP could employ another suitable symbol, alphabet etc.. Nice way +1
    – mrs
    Aug 23 at 8:03












up vote
1
down vote



accepted







up vote
1
down vote



accepted






The spurious solution is introduced by the squaring. Before squaring, $r-2x$ is constrained to have the sign of $o$ (as the square root is a positive number). When squaring, you drop this condition.



Alternative resolution:



Let $z:=sqrtr-x>0$.



The equation becomes



$$z-fracr-z^2z-o=0$$



or



$$2z^2-oz-r=0.$$



The only positive solutions in $z$ is



$$z=fracocolorgreen+sqrto^2+8r4.$$



It exist iff $o^2+8rge0$, and $x=r-z^2$ is implicitly $le r$.






share|cite|improve this answer














The spurious solution is introduced by the squaring. Before squaring, $r-2x$ is constrained to have the sign of $o$ (as the square root is a positive number). When squaring, you drop this condition.



Alternative resolution:



Let $z:=sqrtr-x>0$.



The equation becomes



$$z-fracr-z^2z-o=0$$



or



$$2z^2-oz-r=0.$$



The only positive solutions in $z$ is



$$z=fracocolorgreen+sqrto^2+8r4.$$



It exist iff $o^2+8rge0$, and $x=r-z^2$ is implicitly $le r$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 23 at 7:49

























answered Aug 23 at 7:38









Yves Daoust

113k665207




113k665207







  • 1




    thank you, this is a very elegant way
    – Fabio Dalla Libera
    Aug 23 at 7:56










  • How bad does that $o$ look there. :-) Indeed, the OP could employ another suitable symbol, alphabet etc.. Nice way +1
    – mrs
    Aug 23 at 8:03












  • 1




    thank you, this is a very elegant way
    – Fabio Dalla Libera
    Aug 23 at 7:56










  • How bad does that $o$ look there. :-) Indeed, the OP could employ another suitable symbol, alphabet etc.. Nice way +1
    – mrs
    Aug 23 at 8:03







1




1




thank you, this is a very elegant way
– Fabio Dalla Libera
Aug 23 at 7:56




thank you, this is a very elegant way
– Fabio Dalla Libera
Aug 23 at 7:56












How bad does that $o$ look there. :-) Indeed, the OP could employ another suitable symbol, alphabet etc.. Nice way +1
– mrs
Aug 23 at 8:03




How bad does that $o$ look there. :-) Indeed, the OP could employ another suitable symbol, alphabet etc.. Nice way +1
– mrs
Aug 23 at 8:03

















 

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