Lösung 3.2:4

Aus Online Mathematik Brückenkurs 1

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Square both sides of the equation so that the root sign disappears,
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<center> [[Image:3_2_4.gif]] </center>
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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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<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.

Version vom 10:37, 25. Sep. 2008

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


\displaystyle 1-x=\left( 2-x \right)^{2}\quad \Leftrightarrow \quad 1-x=4-4x+x^{2}


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


\displaystyle \begin{align} & x^{2}-3x+3=0 \\ & \left( x-\frac{3}{2} \right)^{2}-\left( \frac{3}{2} \right)^{2}+3=0 \\ & \left( x-\frac{3}{2} \right)^{2}-\frac{9}{4}+\frac{12}{4}=0 \\ & \left( x-\frac{3}{2} \right)^{2}+\frac{3}{4}=0 \\ \end{align}


As can be seen, the second-order equation does not have any solutions (the left-hand side is always greater than or equal to \displaystyle {3}/{4}\;, regardless of how \displaystyle 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“