Lösung 2.3:4b
Aus Online Mathematik Brückenkurs 1
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| Zeile 1: | Zeile 1: | ||
A first-degree equation which has <math>x=1+\sqrt{3}</math> as a root is <math>x-(1+\sqrt{3}\,)=0</math>, which we can also write as <math>x-1-\sqrt{3} = 0</math>. In the same way, we have that <math>x-(1-\sqrt{3}\,)=0</math>, i.e., <math>x-1+\sqrt{3}=0</math> is a first-degree equation that has <math>x=1-\sqrt{3}</math> as a root. If we multiply these two first-degree equations together, we get a second-degree equation with <math>x=1+\sqrt{3}</math> and <math>x=1-\sqrt{3}</math> as roots, | A first-degree equation which has <math>x=1+\sqrt{3}</math> as a root is <math>x-(1+\sqrt{3}\,)=0</math>, which we can also write as <math>x-1-\sqrt{3} = 0</math>. In the same way, we have that <math>x-(1-\sqrt{3}\,)=0</math>, i.e., <math>x-1+\sqrt{3}=0</math> is a first-degree equation that has <math>x=1-\sqrt{3}</math> as a root. If we multiply these two first-degree equations together, we get a second-degree equation with <math>x=1+\sqrt{3}</math> and <math>x=1-\sqrt{3}</math> as roots, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>(x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,) = 0\,\textrm{.}</math>}} |
The first factor become zero when <math>x=1+\sqrt{3}</math> and the second factor becomes zero when <math>x=1-\sqrt{3}\,</math>. | The first factor become zero when <math>x=1+\sqrt{3}</math> and the second factor becomes zero when <math>x=1-\sqrt{3}\,</math>. | ||
| Zeile 7: | Zeile 7: | ||
Nothing really prevents us from answering with <math>(x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,) = 0</math>, but if we want to give the equation in standard form, we need to expand the left-hand side, | Nothing really prevents us from answering with <math>(x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,) = 0</math>, but if we want to give the equation in standard form, we need to expand the left-hand side, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
(x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,) | (x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,) | ||
&= x^{2} - x + \sqrt{3}x - x + 1 - \sqrt{3} - \sqrt{3}x + \sqrt{3} - (\sqrt{3}\,)^{2}\\[5pt] | &= x^{2} - x + \sqrt{3}x - x + 1 - \sqrt{3} - \sqrt{3}x + \sqrt{3} - (\sqrt{3}\,)^{2}\\[5pt] | ||
| Zeile 19: | Zeile 19: | ||
Note: Exactly as in exercise a, we can multiply the equation by a non-zero constant ''a'' | Note: Exactly as in exercise a, we can multiply the equation by a non-zero constant ''a'' | ||
| - | {{ | + | {{Abgesetzte Formel||<math>ax^{2}-2ax-2a=0</math>}} |
and still have a second-degree equation with the same roots. | and still have a second-degree equation with the same roots. | ||
Version vom 08:33, 22. Okt. 2008
A first-degree equation which has \displaystyle x=1+\sqrt{3} as a root is \displaystyle x-(1+\sqrt{3}\,)=0, which we can also write as \displaystyle x-1-\sqrt{3} = 0. In the same way, we have that \displaystyle x-(1-\sqrt{3}\,)=0, i.e., \displaystyle x-1+\sqrt{3}=0 is a first-degree equation that has \displaystyle x=1-\sqrt{3} as a root. If we multiply these two first-degree equations together, we get a second-degree equation with \displaystyle x=1+\sqrt{3} and \displaystyle x=1-\sqrt{3} as roots,
| \displaystyle (x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,) = 0\,\textrm{.} |
The first factor become zero when \displaystyle x=1+\sqrt{3} and the second factor becomes zero when \displaystyle x=1-\sqrt{3}\,.
Nothing really prevents us from answering with \displaystyle (x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,) = 0, but if we want to give the equation in standard form, we need to expand the left-hand side,
| \displaystyle \begin{align}
(x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,) &= x^{2} - x + \sqrt{3}x - x + 1 - \sqrt{3} - \sqrt{3}x + \sqrt{3} - (\sqrt{3}\,)^{2}\\[5pt] &= x^{2} + (-x+\sqrt{3}x-x-\sqrt{3}x) + (1-\sqrt{3}+\sqrt{3}-3)\\[5pt] &= x^{2}-2x-2 \end{align} |
to get the equation \displaystyle x^{2}-2x-2=0\,.
Note: Exactly as in exercise a, we can multiply the equation by a non-zero constant a
| \displaystyle ax^{2}-2ax-2a=0 |
and still have a second-degree equation with the same roots.
