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Zuletzt geändert von akukin am 2025/07/10 10:07
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akukin 1.1 1 **Zwei Lösungen: D > 0**
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3 1.a) {{formula}}x_1 = -2; \ x_2 = 0,5{{/formula}} {{formula}}\text{L} =\{-2; 0,5\}{{/formula}}
4 1.b) {{formula}}x_1 = -3; \ x_2 = 1{{/formula}} {{formula}}\text{L} =\{-3; 1\}{{/formula}}
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6 **Eine Lösung: D = 0**
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8 2.a) {{formula}}x_{1,2} = 6{{/formula}} {{formula}}\text{L} = \{6\}{{/formula}}
9 2.b) {{formula}}x_{1,2} = 5{{/formula}} {{formula}}\text{L} =\{5\}{{/formula}}
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11 **Keine Lösung: D < 0**
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13 3.a) {{formula}}\text{L} = \emptyset{{/formula}}
14 3.b) {{formula}}\text{L} = \emptyset{{/formula}}
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17 **Effektivstes Verfahren**
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19 4.a) {{formula}}x_1 = -\sqrt{12} \, (= -2\sqrt{3}); \, x_2 = \sqrt{12} \, (= 2\sqrt{3}){{/formula}} **Wurzel ziehen**
20 {{formula}}\text{L} = \{-\sqrt{12}; \sqrt{12}\}{{/formula}}
21 4.b) {{formula}}x_1 = -3; \ x_2 = 3{{/formula}} {{formula}}\text{L} = \{-3; 3\}{{/formula}}
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25 5.a) {{formula}}x_1 = -0,75; \, x_2 = 0,5{{/formula}} {{formula}}\text{L} = \{-0,75; \, 0,5\}{{/formula}} **Satz vom Nullprodukt**
26 5.b) {{formula}}x_1 = -2; \, x_2 = 6{{/formula}} {{formula}}\text{L} = \{-2; \, 6\}{{/formula}}
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29 6.a) {{formula}}x_1 = 0; \, x_2 = 1,5{{/formula}} {{formula}}\text{L} = \{0; \, 1,5\}{{/formula}} **Ausklammern und Satz vom Nullprodukt**
30 6.b) {{formula}}x_1 = 0,2; \, x_2 = 0{{/formula}} {{formula}}\text{L} = \{0; \, 0,2\}{{/formula}}
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