# 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).

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 |