Lösung 3.4:6
Aus Online Mathematik Brückenkurs 2
K (Lösning 3.4:6 moved to Solution 3.4:6: Robot: moved page) |
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| - | {{ | + | First, we try to determine the pure imaginary root. |
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| - | {{ | + | We can write the imaginary root as , where is a real number. If we substitute in , the equation should then be satisfied, |
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| + | <math>\left( ia \right)^{4}+3\left( ia \right)^{3}+\left( ia \right)^{2}+18\left( ia \right)-30=0</math> | ||
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| + | i.e. | ||
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| + | <math>a^{4}-3^{3}i-a^{2}+18ai-30=0</math> | ||
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| + | and, if collect together the real and imaginary parts on the left-hand side, we have | ||
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| + | <math>\left( a^{4}-a^{2}-30 \right)+a\left( -3a^{2}+18 \right)i=0</math> | ||
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| + | If both sides are to be equal, the left-hand side's real and imaginary parts must be zero, | ||
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| + | <math>\left\{ \begin{array}{*{35}l} | ||
| + | a^{4}-a^{2}-30=0 \\ | ||
| + | a\left( -3a^{2}+18 \right)=0 \\ | ||
| + | \end{array} \right.</math> | ||
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| + | The other relation gives | ||
| + | <math>a=0\text{ }</math> | ||
| + | or | ||
| + | <math>a=\pm \sqrt{6}</math>, but it is only | ||
| + | <math>a=\pm \sqrt{6}</math> | ||
| + | which satisfies the first relation. | ||
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| + | Thus, the equation | ||
| + | <math>z^{4}+3z^{3}+z^{2}+18z-30=0</math> | ||
| + | has two pure imaginary roots, | ||
| + | <math>z=-i\sqrt{6}</math> | ||
| + | and | ||
| + | <math>z=i\sqrt{6}</math>. Note that it is completely normal to obtain two imaginary roots. | ||
| + | The polynomial equation has real coefficients and must therefore have complex conjugate roots. | ||
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| + | Now we tackle the problem of determining the equation's other two roots. Because we know that the equation has the two roots | ||
| + | <math>z=\pm i\sqrt{6}</math>, the factor theorem gives that the equation contains the factor | ||
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| + | <math>\left( z-i\sqrt{6} \right)\left( z+i\sqrt{6} \right)=z^{2}+6</math> | ||
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| + | i.e. we can factorize the left-hand side of the equation in the following way, | ||
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| + | <math>z^{4}+3z^{3}+z^{2}+18z-30=\left( z^{2}+Az+B \right)\left( z^{2}+6 \right)</math>, | ||
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| + | where the equation's two other roots are zeros of the unknown factor | ||
| + | <math>z^{2}+Az+B</math>. | ||
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| + | We determine the factor | ||
| + | <math>z^{2}+Az+B</math> | ||
| + | by means of a polynomial division (divide both sides by | ||
| + | <math>z^{2}+6</math>): | ||
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| + | <math>\begin{align} | ||
| + | & z^{2}+Az+B=\frac{z^{4}+3z^{3}+z^{2}+18z-30}{z^{2}+6} \\ | ||
| + | & =\frac{z^{4}+6z^{2}-6z^{2}+3z^{3}+z^{2}+18z-30}{z^{2}+6} \\ | ||
| + | & =\frac{z^{2}\left( z^{2}+6 \right)+3z^{3}-5z^{2}+18z-30}{z^{2}+6} \\ | ||
| + | & =z^{2}+\frac{3z^{3}-5z^{2}+18z-30}{z^{2}+6} \\ | ||
| + | & =z^{2}+\frac{3z^{3}+18z-18z-5z^{2}+18z-30}{z^{2}+6} \\ | ||
| + | & =z^{2}+\frac{3z\left( z^{2}+6 \right)-5z^{2}-30}{z^{2}+6} \\ | ||
| + | & =z^{2}+3z+\frac{-5z^{2}-30}{z^{2}+6} \\ | ||
| + | & =z^{2}+3z+\frac{-5\left( z^{2}+6 \right)}{z^{2}+6} \\ | ||
| + | & =z^{2}+3z-5 \\ | ||
| + | \end{align}</math> | ||
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| + | To obtain the two remaining roots, we need therefore to solve the equation | ||
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| + | <math>z^{2}+3z-5=0</math> | ||
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| + | We complete the square | ||
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| + | <math>\begin{align} | ||
| + | & \left( z+\frac{3}{2} \right)^{2}-\left( \frac{3}{2} \right)^{2}-5=0 \\ | ||
| + | & \left( z+\frac{3}{2} \right)^{2}=\frac{29}{4} \\ | ||
| + | \end{align}</math> | ||
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| + | which gives that | ||
| + | <math>z=-\frac{3}{2}\pm \frac{\sqrt{29}}{2}</math>. | ||
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| + | The answer is that the equation has the roots | ||
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| + | <math>z=-i\sqrt{6},\ z=i\sqrt{6},\ z=-\frac{3}{2}-\frac{\sqrt{29}}{2},\ z=-\frac{3}{2}+\frac{\sqrt{29}}{2}</math> | ||
Version vom 14:02, 28. Okt. 2008
First, we try to determine the pure imaginary root.
We can write the imaginary root as , where is a real number. If we substitute in , the equation should then be satisfied,
\displaystyle \left( ia \right)^{4}+3\left( ia \right)^{3}+\left( ia \right)^{2}+18\left( ia \right)-30=0
i.e.
\displaystyle a^{4}-3^{3}i-a^{2}+18ai-30=0
and, if collect together the real and imaginary parts on the left-hand side, we have
\displaystyle \left( a^{4}-a^{2}-30 \right)+a\left( -3a^{2}+18 \right)i=0
If both sides are to be equal, the left-hand side's real and imaginary parts must be zero,
\displaystyle \left\{ \begin{array}{*{35}l}
a^{4}-a^{2}-30=0 \\
a\left( -3a^{2}+18 \right)=0 \\
\end{array} \right.
The other relation gives
\displaystyle a=0\text{ }
or
\displaystyle a=\pm \sqrt{6}, but it is only
\displaystyle a=\pm \sqrt{6}
which satisfies the first relation.
Thus, the equation \displaystyle z^{4}+3z^{3}+z^{2}+18z-30=0 has two pure imaginary roots, \displaystyle z=-i\sqrt{6} and \displaystyle z=i\sqrt{6}. Note that it is completely normal to obtain two imaginary roots. The polynomial equation has real coefficients and must therefore have complex conjugate roots.
Now we tackle the problem of determining the equation's other two roots. Because we know that the equation has the two roots \displaystyle z=\pm i\sqrt{6}, the factor theorem gives that the equation contains the factor
\displaystyle \left( z-i\sqrt{6} \right)\left( z+i\sqrt{6} \right)=z^{2}+6
i.e. we can factorize the left-hand side of the equation in the following way,
\displaystyle z^{4}+3z^{3}+z^{2}+18z-30=\left( z^{2}+Az+B \right)\left( z^{2}+6 \right),
where the equation's two other roots are zeros of the unknown factor \displaystyle z^{2}+Az+B.
We determine the factor \displaystyle z^{2}+Az+B by means of a polynomial division (divide both sides by \displaystyle z^{2}+6):
\displaystyle \begin{align}
& z^{2}+Az+B=\frac{z^{4}+3z^{3}+z^{2}+18z-30}{z^{2}+6} \\
& =\frac{z^{4}+6z^{2}-6z^{2}+3z^{3}+z^{2}+18z-30}{z^{2}+6} \\
& =\frac{z^{2}\left( z^{2}+6 \right)+3z^{3}-5z^{2}+18z-30}{z^{2}+6} \\
& =z^{2}+\frac{3z^{3}-5z^{2}+18z-30}{z^{2}+6} \\
& =z^{2}+\frac{3z^{3}+18z-18z-5z^{2}+18z-30}{z^{2}+6} \\
& =z^{2}+\frac{3z\left( z^{2}+6 \right)-5z^{2}-30}{z^{2}+6} \\
& =z^{2}+3z+\frac{-5z^{2}-30}{z^{2}+6} \\
& =z^{2}+3z+\frac{-5\left( z^{2}+6 \right)}{z^{2}+6} \\
& =z^{2}+3z-5 \\
\end{align}
To obtain the two remaining roots, we need therefore to solve the equation
\displaystyle z^{2}+3z-5=0
We complete the square
\displaystyle \begin{align}
& \left( z+\frac{3}{2} \right)^{2}-\left( \frac{3}{2} \right)^{2}-5=0 \\
& \left( z+\frac{3}{2} \right)^{2}=\frac{29}{4} \\
\end{align}
which gives that
\displaystyle z=-\frac{3}{2}\pm \frac{\sqrt{29}}{2}.
The answer is that the equation has the roots
\displaystyle z=-i\sqrt{6},\ z=i\sqrt{6},\ z=-\frac{3}{2}-\frac{\sqrt{29}}{2},\ z=-\frac{3}{2}+\frac{\sqrt{29}}{2}
