Wiki-Quellcode von Lösung Lösen
Version 5.1 von Dirk Tebbe am 2026/03/27 10:08
Verstecke letzte Bearbeiter
| author | version | line-number | content |
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1.1 | 1 | a) {{formula}}0= -x^3 -4096 \Rightarrow x^3= -4096 \Rightarrow x = \sqrt[3]{-4096} = -16{{/formula}} |
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1.2 | 3 | b) {{formula}}0= x^{2}(x+3)(x-3)(x+8){{/formula}} mit dem Satz vom Nullprodukt folgt {{formula}}x_{1,2} = 0 \lor x_3 =- 3 \lor x_4 = 3 \lor x_5 = 8{{/formula}} |
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1.1 | 4 | |
| 5 | c) {{formula}}0= x^4-2x^2-35{{/formula}} | ||
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3.1 | 6 | Substitution {{formula}} x^2 = z \Rightarrow 0= z^2 -2z-35 {{/formula}} |
| 7 | Lösung mit der a,b,c-Formel: {{formula}}z_{1,2} = \frac{2 \pm \sqrt{(-2)^2-4 \cdot 1 \cdot (-35) } }{2} \Rightarrow z_1 = 7 ; z_2 = -5{{/formula}} | ||
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4.1 | 8 | Rücksubstitution: |
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3.1 | 9 | {{formula}} z = x^2 = 7 \Rightarrow x_{1,2}= \pm \sqrt{7} {{/formula}} |
| 10 | {{formula}} z = x^2 = -5 {{/formula}} liefert keine weitere Lösung. | ||
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1.1 | 11 | |
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5.1 | 12 | d) {{formula}}x_1 = -2 \lor x_2 = 2 \lor x_3 = 3 {{/formula}} |
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