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Homestrange root

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# strange root

In solving certain types of equations, one may obtain besides the proper (right) roots 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 current 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 (but the equations $a=\pm b$ and $a^{2}=b^{2}$ would be such ones).

## Mathematics Subject Classification

97D99*no label found*26A09

*no label found*

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