Wiki-Quellcode von Lösung Rückwärts lösen
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| author | version | line-number | content |
|---|---|---|---|
| 1 | a) Aus {{formula}}x=-2{{/formula}} ergibt sich durch Potenzieren mit {{formula}}3{{/formula}}: {{formula}}x^3=-8{{/formula}} | ||
| 2 | |||
| 3 | Multiplizieren mit {{formula}}2{{/formula}} ergibt {{formula}}2x^3=-16{{/formula}} | ||
| 4 | |||
| 5 | Addieren von {{formula}}16{{/formula}} auf beiden Seiten ergibt {{formula}}2x^3+16=0{{/formula}} | ||
| 6 | |||
| 7 | Insgesamt folgt also: | ||
| 8 | |||
| 9 | {{formula}} | ||
| 10 | \begin{align} | ||
| 11 | 2x^3+16&=0 \\ | ||
| 12 | 2x^3&=-16 \quad \mid :2 \\ | ||
| 13 | x^3&=-8 \\ | ||
| 14 | x&=-2 | ||
| 15 | \end{align} | ||
| 16 | {{/formula}} | ||
| 17 | |||
| 18 | b) Die Gleichung hat die Lösungen {{formula}}x_{1,2}=0{{/formula}} und {{formula}}x_3=6{{/formula}}. In der Produktform/Nullstellenform ergibt sich: | ||
| 19 | |||
| 20 | {{formula}} | ||
| 21 | \begin{align} | ||
| 22 | 2x^3(x-6)=0 \\ | ||
| 23 | \Leftrightarrow 2x^3-12x^2=0 | ||
| 24 | \end{align} | ||
| 25 | {{/formula}} | ||
| 26 | |||
| 27 | c) {{formula}}(\pm 2)^2=4{{/formula}} | ||
| 28 | Damit ergeben sich die beiden Lösungen {{formula}}z_1=36{{/formula}} und {{formula}}z_2=4{{/formula}}. In der Produktform/Nullstellenform ergibt sich {{formula}}(z-36)(z-4)=0{{/formula}}. Ausmultiplizieren führt auf {{formula}}z^2-4z-36z+144=0{{/formula}} | ||
| 29 | |||
| 30 | {{formula}} | ||
| 31 | \begin{align} | ||
| 32 | z^2-4z-36z+144=0 \\ | ||
| 33 | \Leftrightarrow z^2-40z+144=0 | ||
| 34 | \end{align} | ||
| 35 | {{/formula}} | ||
| 36 | |||
| 37 | Insgesamt ergibt sich {{formula}}x^4-40x^2+144=0{{/formula}} | ||
| 38 | |||
| 39 | {{formula}}\begin{align*} | ||
| 40 | x^4-40x^2+144 &= 0 & \left|\left|\text{ Subst.: } & x^2:=z\\ | ||
| 41 | z^2-40z+144=0 &= 0 & \left|\left|\text{ SVNP } & | ||
| 42 | \end{align*} | ||
| 43 | {{/formula}} | ||
| 44 | |||
| 45 | {{formula}} | ||
| 46 | \begin{align*} | ||
| 47 | \Rightarrowz_{1,2}&=\frac{\square\pm\sqrt{\square^2-4\cdot\square\cdot\square}}{2\cdot\square}\\ | ||
| 48 | z_1&=\frac{\square+\square}{\square}; z_2=\frac{\square-\square}{\square} | ||
| 49 | \end{align*} | ||
| 50 | {{/formula}} | ||
| 51 | |||
| 52 | {{formula}} | ||
| 53 | \begin{align*} | ||
| 54 | &\text{Resubst.: } \square := x^2\\ | ||
| 55 | &x_{1,2}^2=36 \Rightarrow x_{1,2}=\square\\ | ||
| 56 | &x_{3,4}^2=\square \Rightarrow x_{3,4}=\pm 2 | ||
| 57 | \end{align*} | ||
| 58 | {{/formula}} | ||
| 59 | |||
| 60 | [[image:Handschriftlich.jpg]] |