Lösung 3.4:6

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 3: Zeile 3:
We can write the imaginary root as <math>z=ia</math>, where <math>a</math> is a real number. If we substitute in <math>z=ia</math>, the equation should then be satisfied,
We can write the imaginary root as <math>z=ia</math>, where <math>a</math> is a real number. If we substitute in <math>z=ia</math>, the equation should then be satisfied,
-
{{Displayed math||<math>(ia)^4 + 3(ia)^3 + (ia)^2 + 18(ia) - 30 = 0\,,</math>}}
+
{{Abgesetzte Formel||<math>(ia)^4 + 3(ia)^3 + (ia)^2 + 18(ia) - 30 = 0\,,</math>}}
i.e.
i.e.
-
{{Displayed math||<math>a^4 - 3a^3i - a^2 + 18ai - 30 = 0</math>}}
+
{{Abgesetzte Formel||<math>a^4 - 3a^3i - a^2 + 18ai - 30 = 0</math>}}
and, if collect together the real and imaginary parts on the left-hand side, we have
and, if collect together the real and imaginary parts on the left-hand side, we have
-
{{Displayed math||<math>(a^4-a^2-30) + a(-3a^2+18)i = 0\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>(a^4-a^2-30) + a(-3a^2+18)i = 0\,\textrm{.}</math>}}
If both sides are to be equal, the left-hand side's real and imaginary parts must be zero,
If both sides are to be equal, the left-hand side's real and imaginary parts must be zero,
-
{{Displayed math||<math>\left\{\begin{align}
+
{{Abgesetzte Formel||<math>\left\{\begin{align}
a^4-a^2-30 &= 0\,,\\[5pt]
a^4-a^2-30 &= 0\,,\\[5pt]
a(-3a^2+18) &= 0\,\textrm{.}
a(-3a^2+18) &= 0\,\textrm{.}
Zeile 26: Zeile 26:
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
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
-
{{Displayed math||<math>(z-i\sqrt{6})(z+i\sqrt{6}) = z^2+6\,,</math>}}
+
{{Abgesetzte Formel||<math>(z-i\sqrt{6})(z+i\sqrt{6}) = z^2+6\,,</math>}}
i.e. we can factorize the left-hand side of the equation in the following way,
i.e. we can factorize the left-hand side of the equation in the following way,
-
{{Displayed math||<math>z^4+3z^3+z^2+18z-30 = (z^2+Az+B)(z^2+6)\,,</math>}}
+
{{Abgesetzte Formel||<math>z^4+3z^3+z^2+18z-30 = (z^2+Az+B)(z^2+6)\,,</math>}}
where the equation's two other roots are zeros of the unknown factor <math>z^{2}+Az+B</math>.
where the equation's two other roots are zeros of the unknown factor <math>z^{2}+Az+B</math>.
Zeile 36: Zeile 36:
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>),
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>),
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
z^2+Az+B
z^2+Az+B
&= \frac{z^4+3z^3+z^2+18z-30}{z^2+6}\\[5pt]
&= \frac{z^4+3z^3+z^2+18z-30}{z^2+6}\\[5pt]
Zeile 51: Zeile 51:
To obtain the two remaining roots, we need therefore to solve the equation
To obtain the two remaining roots, we need therefore to solve the equation
-
{{Displayed math||<math>z^2+3z-5 = 0\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>z^2+3z-5 = 0\,\textrm{.}</math>}}
We complete the square
We complete the square
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\Bigl(z+\frac{3}{2}\Bigr)^2 - \Bigl(\frac{3}{2}\Bigr)^2 - 5 &= 0\,,\\[5pt]
\Bigl(z+\frac{3}{2}\Bigr)^2 - \Bigl(\frac{3}{2}\Bigr)^2 - 5 &= 0\,,\\[5pt]
\Bigl(z+\frac{3}{2}\Bigr)^2 &= \frac{29}{4}\,,
\Bigl(z+\frac{3}{2}\Bigr)^2 &= \frac{29}{4}\,,
Zeile 64: Zeile 64:
The answer is that the equation has the roots
The answer is that the equation has the roots
-
{{Displayed math||<math>z=-i\sqrt{6}</math>, <math>\quad z=i\sqrt{6}</math>, <math>\quad z=-\frac{3}{2}-\frac{\sqrt{29}}{2}</math>, <math>\quad z=-\frac{3}{2}+\frac{\sqrt{29}}{2}\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>z=-i\sqrt{6}</math>, <math>\quad z=i\sqrt{6}</math>, <math>\quad z=-\frac{3}{2}-\frac{\sqrt{29}}{2}</math>, <math>\quad z=-\frac{3}{2}+\frac{\sqrt{29}}{2}\,\textrm{.}</math>}}

Version vom 13:16, 10. Mär. 2009

First, we try to determine the pure imaginary root.

We can write the imaginary root as \displaystyle z=ia, where \displaystyle a is a real number. If we substitute in \displaystyle z=ia, the equation should then be satisfied,

\displaystyle (ia)^4 + 3(ia)^3 + (ia)^2 + 18(ia) - 30 = 0\,,

i.e.

\displaystyle a^4 - 3a^3i - a^2 + 18ai - 30 = 0

and, if collect together the real and imaginary parts on the left-hand side, we have

\displaystyle (a^4-a^2-30) + a(-3a^2+18)i = 0\,\textrm{.}

If both sides are to be equal, the left-hand side's real and imaginary parts must be zero,

\displaystyle \left\{\begin{align}

a^4-a^2-30 &= 0\,,\\[5pt] a(-3a^2+18) &= 0\,\textrm{.} \end{align}\right.

The other relation gives \displaystyle a=0 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 (z-i\sqrt{6})(z+i\sqrt{6}) = 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 = (z^2+Az+B)(z^2+6)\,,

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}\\[5pt] &= \frac{z^4+6z^2-6z^2+3z^3+z^2+18z-30}{z^2+6}\\[5pt] &= \frac{z^2(z^2+6)+3z^3-5z^2+18z-30}{z^2+6}\\[5pt] &= z^2 + \frac{3z^3-5z^2+18z-30}{z^2+6}\\[5pt] &= z^2 + \frac{3z^3+18z-18z-5z^2+18z-30}{z^2+6}\\[5pt] &= z^2 + \frac{3z(z^2+6)-5z^2-30}{z^2+6}\\[5pt] &= z^2 + 3z + \frac{-5z^2-30}{z^2+6}\\[5pt] &= z^2 + 3z + \frac{-5(z^2+6)}{z^2+6}\\[5pt] &= z^2 + 3z - 5\,\textrm{.} \end{align}

To obtain the two remaining roots, we need therefore to solve the equation

\displaystyle z^2+3z-5 = 0\,\textrm{.}

We complete the square

\displaystyle \begin{align}

\Bigl(z+\frac{3}{2}\Bigr)^2 - \Bigl(\frac{3}{2}\Bigr)^2 - 5 &= 0\,,\\[5pt] \Bigl(z+\frac{3}{2}\Bigr)^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}, \displaystyle \quad z=i\sqrt{6}, \displaystyle \quad z=-\frac{3}{2}-\frac{\sqrt{29}}{2}, \displaystyle \quad z=-\frac{3}{2}+\frac{\sqrt{29}}{2}\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.4:6“