Lösung 3.2:4

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Square both sides of the equation so that the root sign disappears,
Square both sides of the equation so that the root sign disappears,
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{{Displayed math||<math>1-x = (2-x)^2\quad \Leftrightarrow \quad 1-x = 4-4x+x^2</math>}}
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<math>1-x=\left( 2-x \right)^{2}\quad \Leftrightarrow \quad 1-x=4-4x+x^{2}</math>
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and then solve the resulting second-order equation by completing the square,
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{{Displayed math||<math>\begin{align}
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x^{2}-3x+3 &= 0\,,\\[5pt]
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\Bigl(x-\frac{3}{2}\Bigr)^{2} - \Bigl(\frac{3}{2}\Bigr)^{2} + 3 &= 0\,,\\[5pt]
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\Bigl(x-\frac{3}{2}\Bigr)^{2} - \frac{9}{4} + \frac{12}{4} &= 0\,,\\[5pt]
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\Bigl(x-\frac{3}{2}\Bigr)^{2} + \frac{3}{4} &= 0\,\textrm{.}
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\end{align}</math>}}
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and then solve the resulting second-order equation by completing the square:
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As can be seen, the second-order equation does not have any solutions (the left-hand side is always greater than or equal to 3/4, regardless of how ''x'' is chosen) so, the original root equation does not have any solutions.
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<math>\begin{align}
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& x^{2}-3x+3=0 \\
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& \left( x-\frac{3}{2} \right)^{2}-\left( \frac{3}{2} \right)^{2}+3=0 \\
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& \left( x-\frac{3}{2} \right)^{2}-\frac{9}{4}+\frac{12}{4}=0 \\
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& \left( x-\frac{3}{2} \right)^{2}+\frac{3}{4}=0 \\
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\end{align}</math>
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As can be seen, the second-order equation does not have any solutions (the left-hand side is always greater than or equal to
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<math>{3}/{4}\;</math>, regardless of how
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<math>x</math>
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is chosen; so, the original root equation does not have any solutions.
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Version vom 12:42, 1. Okt. 2008

Square both sides of the equation so that the root sign disappears,

Vorlage:Displayed math

and then solve the resulting second-order equation by completing the square,

Vorlage:Displayed math

As can be seen, the second-order equation does not have any solutions (the left-hand side is always greater than or equal to 3/4, regardless of how x is chosen) so, the original root equation does not have any solutions.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.2:4“