Lösung 2.3:4b

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A first-degree equation which
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<math>x=\text{ 1}+\sqrt{\text{3}}</math>
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as a root is
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<math>x-\left( \text{1}+\sqrt{\text{3}} \right)=0</math>, which we can also write as
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<math>x-\text{1-}\sqrt{\text{3}}=0</math>. In the same way, we have that
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<math>x-\left( \text{1-}\sqrt{\text{3}} \right)=0</math>, i.e.,
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<math>x-\text{1+}\sqrt{\text{3}}=0</math>
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is a first-degree equation that has
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<math>x=\text{ 1-}\sqrt{\text{3}}</math>
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as a root. If we multiply these two first-degree equations together, we get a second-degree equation with
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<math>x=\text{ 1}+\sqrt{\text{3}}</math>
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and
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<math>x=\text{ 1-}\sqrt{\text{3}}</math>. as roots:
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<math>\left( x-\text{1-}\sqrt{\text{3}} \right)\left( x-\text{1+}\sqrt{\text{3}} \right)=0</math>
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The first factor become zero when
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<math>x=\text{ 1}+\sqrt{\text{3}}</math>
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and the second factor becomes zero when
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<math>x=\text{ 1-}\sqrt{\text{3}}</math>.
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Nothing really prevents us from answering with
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<math>\left( x-\text{1-}\sqrt{\text{3}} \right)\left( x-\text{1+}\sqrt{\text{3}} \right)=0</math>, but if we want to give the equation in standard form, we need to expand the left-hand side,
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<math>\begin{align}
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& \left( x-\text{1-}\sqrt{\text{3}} \right)\left( x-\text{1+}\sqrt{\text{3}} \right)=x^{2}-x+\sqrt{3}-x+1-\sqrt{3}-\sqrt{3}x+\sqrt{3}-\left( \sqrt{3} \right)^{2} \\
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& =x^{2}+\left( -x+\sqrt{3}x-x-\sqrt{3}x \right)+\left( 1-\sqrt{3}+\sqrt{3}-3 \right) \\
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& =x^{2}-2x-2 \\
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\end{align}</math>
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to get the equation
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<math>x^{2}-2x-2=0</math>
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NOTE: Exactly as in exercise
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<math>\text{a}</math>, we can multiply the equation by a non-zero constant
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<math>a</math>,
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<math>ax^{2}-2ax-2a=0</math>
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and still have a second-degree equation with the same roots.

Version vom 09:29, 21. Sep. 2008

A first-degree equation which \displaystyle x=\text{ 1}+\sqrt{\text{3}} as a root is \displaystyle x-\left( \text{1}+\sqrt{\text{3}} \right)=0, which we can also write as

\displaystyle x-\text{1-}\sqrt{\text{3}}=0. In the same way, we have that \displaystyle x-\left( \text{1-}\sqrt{\text{3}} \right)=0, i.e., \displaystyle x-\text{1+}\sqrt{\text{3}}=0 is a first-degree equation that has \displaystyle x=\text{ 1-}\sqrt{\text{3}} as a root. If we multiply these two first-degree equations together, we get a second-degree equation with \displaystyle x=\text{ 1}+\sqrt{\text{3}} and \displaystyle x=\text{ 1-}\sqrt{\text{3}}. as roots:


\displaystyle \left( x-\text{1-}\sqrt{\text{3}} \right)\left( x-\text{1+}\sqrt{\text{3}} \right)=0


The first factor become zero when \displaystyle x=\text{ 1}+\sqrt{\text{3}} and the second factor becomes zero when \displaystyle x=\text{ 1-}\sqrt{\text{3}}.

Nothing really prevents us from answering with \displaystyle \left( x-\text{1-}\sqrt{\text{3}} \right)\left( x-\text{1+}\sqrt{\text{3}} \right)=0, but if we want to give the equation in standard form, we need to expand the left-hand side,


\displaystyle \begin{align} & \left( x-\text{1-}\sqrt{\text{3}} \right)\left( x-\text{1+}\sqrt{\text{3}} \right)=x^{2}-x+\sqrt{3}-x+1-\sqrt{3}-\sqrt{3}x+\sqrt{3}-\left( \sqrt{3} \right)^{2} \\ & =x^{2}+\left( -x+\sqrt{3}x-x-\sqrt{3}x \right)+\left( 1-\sqrt{3}+\sqrt{3}-3 \right) \\ & =x^{2}-2x-2 \\ \end{align}


to get the equation \displaystyle x^{2}-2x-2=0


NOTE: Exactly as in exercise \displaystyle \text{a}, we can multiply the equation by a non-zero constant \displaystyle a,


\displaystyle ax^{2}-2ax-2a=0


and still have a second-degree equation with the same roots.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.3:4b“