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Revision difference : strange root |
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Version 3 |
| In solving certain \PMlinkescapetext{types} of equations, one may obtain besides the proper (\PMlinkescapetext{right}) roots also some {\em 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.\\ |
In solving certain \PMlinkescapetext{types} of equations, one may obtain besides the proper (\PMlinkescapetext{right}) roots also some {\em 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.\\ |
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| \textbf{Example.}\, $$x-\sqrt{x} = 12$$ |
\textbf{Example.}\, $$x-\sqrt{x} = 12$$ |
| $$x-12 = \sqrt{x}$$ |
$$x-12 = \sqrt{x}$$ |
| $$(x-12)^2 = (\sqrt{x})^2$$ |
$$(x-12)^2 = (\sqrt{x})^2$$ |
| $$x^2-24x+144 = x$$ |
$$x^2-24x+144 = x$$ |
| $$x^2-25x+144 = 0$$ |
$$x^2-25x+144 = 0$$ |
| $$x = \frac{25\pm\sqrt{25^2-4\cdot144}}{2} = \frac{25\pm7}{2}$$ |
$$x = \frac{25\pm\sqrt{25^2-4\cdot144}}{2} = \frac{25\pm7}{2}$$ |
| $$x = 16 \quad \lor \quad x = 9$$ |
$$x = 16 \quad \lor \quad x = 9$$ |
| Substituting these values of $x$ into the left side of the original equation yields |
Substituting these values of $x$ into the left side of the original equation yields |
| $$16-4 = 12, \quad 9-3 = 6.$$ |
$$16-4 = 12, \quad 9-3 = 6.$$ |
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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\, $\sqrt{x} = \frac{1\pm7}{2}$,\, 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 \PMlinkescapetext{current} practice excludes them.)\\
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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\, $\sqrt{x} = \frac{1\pm7}{2}$,\, 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.)\\
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| The general explanation of strange roots when squaring an equation is, that the two equations |
The general explanation of strange roots when squaring an equation is, that the two equations |
| $$a = b,$$ |
$$a = b,$$ |
| $$a^2 = b^2$$ |
$$a^2 = b^2$$ |
| are not \PMlinkname{equivalent}{Equivalent3} (but the equations\, $a = \pm b$\, and\, $a^2 = b^2$\, would be such ones). |
are not \PMlinkname{equivalent}{Equivalent3} (but the equations\, $a = \pm b$\, and\, $a^2 = b^2$\, would be such ones). |
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