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.

$$x-\sqrt{x}=12$$

$$x-12=\sqrt{x}$$

$${(x-12)}^2={(\sqrt{x})}^2$$

$$x^2-24x+144=x$$

$$x^2-25x+144=0$$

$$x=\frac{25\pm \sqrt{{25}^2-4\cdot 144}}{2}=\frac{25\pm 7}{2}$$

$$x=16\mathit{\hspace{1em}}\vee \mathit{\hspace{1em}}x=9$$

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

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

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).