# strange root

In solving certain of equations, one may obtain besides the proper () roots (http://planetmath.org/Equation) also some strange roots which do not satisfy the original equation. Such a thing can happen especially when one has in some stage squared both sides of the treated equation; in this situation one must check all “roots” by substituting them to the original equation.

Example.

 $x-\sqrt{x}=12$
 $x-12=\sqrt{x}$
 $(x-12)^{2}=(\sqrt{x})^{2}$
 $x^{2}-24x+144=x$
 $x^{2}-25x+144=0$
 $x=\frac{25\pm\sqrt{25^{2}-4\cdot 144}}{2}=\frac{25\pm 7}{2}$
 $x=16\quad\lor\quad x=9$

Substituting these values of $x$ into the left side of the original equation yields

 $16-4=12,\quad 9-3=6.$

Thus, only  $x=16$  is valid,  $x=9$  is a strange root. (How  $x=9$  is related to the solved equation, is explained by that it may be written  $(\sqrt{x})^{2}-\sqrt{x}-12=0$, from which one would obtain via the quadratic formula that  $\sqrt{x}=\frac{1\pm 7}{2}$,  i.e.  $\sqrt{x}=4$  or  $\sqrt{x}=-3$.  The latter corresponds the value  $x=9$,  but it were relevant to the original equation only if we would allow negative values for square roots of positive numbers; the practice excludes them.)

The general explanation of strange roots when squaring an equation is, that the two equations

 $a=b,$
 $a^{2}=b^{2}$

are not equivalent (http://planetmath.org/Equivalent3) (but the equations  $a=\pm b$  and  $a^{2}=b^{2}$  would be such ones).

 Title strange root Canonical name StrangeRoot Date of creation 2013-03-22 17:55:53 Last modified on 2013-03-22 17:55:53 Owner pahio (2872) Last modified by pahio (2872) Numerical id 8 Author pahio (2872) Entry type Definition Classification msc 97D99 Classification msc 26A09 Synonym wrong root Synonym extraneous root Related topic QuadraticFormula Related topic LogicalOr Related topic SquaringConditionForSquareRootInequality