Lösung 3.2:5

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K (Lösning 3.2:5 moved to Solution 3.2:5: Robot: moved page)
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After squaring both sides, we obtain the equation
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<center> [[Image:3_2_5-1(2).gif]] </center>
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<math>3x-2=\left( 2-x \right)^{2}\quad \quad (*)</math>
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<center> [[Image:3_2_5-2(2).gif]] </center>
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and if we expand the right-hand side and then collect gather the terms, we get
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<math>x^{2}-7x+6=0</math>
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Completing the square of the left-hand side, we obtain
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<math>\begin{align}
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& x^{2}-7x+6=\left( x-\frac{7}{2} \right)^{2}-\left( \frac{7}{2} \right)^{2}+6 \\
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& =\left( x-\frac{7}{2} \right)^{2}-\frac{49}{4}+\frac{24}{4} \\
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& =\left( x-\frac{7}{2} \right)^{2}-\frac{25}{4} \\
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\end{align}</math>
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which means that the equation can be written as
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<math>\left( x-\frac{7}{2} \right)^{2}=\frac{25}{4}</math>
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and the solutions are therefore
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<math>\begin{align}
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& x=\frac{7}{2}+\sqrt{\frac{25}{4}=}\frac{7}{2}+\frac{5}{2}=\frac{12}{2}=6 \\
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& x=\frac{7}{2}-\sqrt{\frac{25}{4}=}\frac{7}{2}-\frac{5}{2}=\frac{2}{2}=1 \\
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\end{align}</math>
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EQ6
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Substituting
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<math>x=\text{1 }</math>
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and
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<math>x=\text{6 }</math>
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into the quadratic equation (*) shows that we have solved the equation correctly.
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<math>x=\text{1 }</math>: LHS
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<math>=3\centerdot 1-2=1</math>
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and RHS
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<math>=\left( 2-1 \right)^{2}=1</math>
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<math>x=\text{6 }</math>: LHS
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<math>=3\centerdot 6-2=16</math>
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and RHS
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<math>=\left( 2-6 \right)^{2}=16</math>
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Finally, we need to sort away possible false roots to the root equation by verifying the solutions.
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<math>x=\text{1 }</math>: LHS
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<math>=\sqrt{3\centerdot 1-2}=1</math>
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and RHS
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<math>=2-1=1</math>
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<math>x=\text{6 }</math>: LHS
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<math>=\sqrt{3\centerdot 6-2}=4</math>
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and RHS =
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<math>=2-6=-4</math>
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This shows that the root equation has the solution
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<math>x=\text{1}</math>.

Version vom 10:54, 25. Sep. 2008

After squaring both sides, we obtain the equation


\displaystyle 3x-2=\left( 2-x \right)^{2}\quad \quad (*)


and if we expand the right-hand side and then collect gather the terms, we get


\displaystyle x^{2}-7x+6=0


Completing the square of the left-hand side, we obtain


\displaystyle \begin{align} & x^{2}-7x+6=\left( x-\frac{7}{2} \right)^{2}-\left( \frac{7}{2} \right)^{2}+6 \\ & =\left( x-\frac{7}{2} \right)^{2}-\frac{49}{4}+\frac{24}{4} \\ & =\left( x-\frac{7}{2} \right)^{2}-\frac{25}{4} \\ \end{align}

which means that the equation can be written as


\displaystyle \left( x-\frac{7}{2} \right)^{2}=\frac{25}{4}


and the solutions are therefore


\displaystyle \begin{align} & x=\frac{7}{2}+\sqrt{\frac{25}{4}=}\frac{7}{2}+\frac{5}{2}=\frac{12}{2}=6 \\ & x=\frac{7}{2}-\sqrt{\frac{25}{4}=}\frac{7}{2}-\frac{5}{2}=\frac{2}{2}=1 \\ \end{align} EQ6

Substituting \displaystyle x=\text{1 } and \displaystyle x=\text{6 } into the quadratic equation (*) shows that we have solved the equation correctly.


\displaystyle x=\text{1 }: LHS \displaystyle =3\centerdot 1-2=1 and RHS \displaystyle =\left( 2-1 \right)^{2}=1


\displaystyle x=\text{6 }: LHS \displaystyle =3\centerdot 6-2=16 and RHS \displaystyle =\left( 2-6 \right)^{2}=16


Finally, we need to sort away possible false roots to the root equation by verifying the solutions.


\displaystyle x=\text{1 }: LHS \displaystyle =\sqrt{3\centerdot 1-2}=1 and RHS \displaystyle =2-1=1


\displaystyle x=\text{6 }: LHS \displaystyle =\sqrt{3\centerdot 6-2}=4 and RHS = \displaystyle =2-6=-4

This shows that the root equation has the solution \displaystyle x=\text{1}.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.2:5“