Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
No jsMath TeX fonts found -- using image fonts instead.
These may be slow and might not print well.
Use the jsMath control panel to get additional information.
jsMath Control PanelHide this Message

jsMath

Lösung 3.4:5

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:
For our equation, this means that the left-hand side can be factorized as
For our equation, this means that the left-hand side can be factorized as
-
{{Displayed math||<math>z^4-6z^2+az+b = (z-c)^3(z-d)</math>}}
+
{{Abgesetzte Formel||<math>z^4-6z^2+az+b = (z-c)^3(z-d)</math>}}
according to the factor theorem, where <math>z=c</math> is the triple root and
according to the factor theorem, where <math>z=c</math> is the triple root and
Zeile 12: Zeile 12:
If we expand the right-hand side above, we get
If we expand the right-hand side above, we get
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
(z-c)^3(z-d)
(z-c)^3(z-d)
&= (z-c)^2(z-c)(z-d)\\[5pt]
&= (z-c)^2(z-c)(z-d)\\[5pt]
Zeile 22: Zeile 22:
and this means that we must have
and this means that we must have
-
{{Displayed math||<math>z^4-6z^2+az+b = z^4-(3c+d)z^3+3c(c+d)z^2-c^2(c-3d)z+c^3d\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>z^4-6z^2+az+b = z^4-(3c+d)z^3+3c(c+d)z^2-c^2(c-3d)z+c^3d\,\textrm{.}</math>}}
Because two polynomials are equal if an only if their coefficients are equal, this gives
Because two polynomials are equal if an only if their coefficients are equal, this gives
-
{{Displayed math||<math>\left\{\begin{align}
+
{{Abgesetzte Formel||<math>\left\{\begin{align}
3c+d &= 0\,,\\[5pt]
3c+d &= 0\,,\\[5pt]
3c(c+d) &= -6\,,\\[5pt]
3c(c+d) &= -6\,,\\[5pt]
Zeile 35: Zeile 35:
From the first equation, we obtain <math>d=-3c</math> and substituting this into the second equation gives us an equation for <math>c</math>,
From the first equation, we obtain <math>d=-3c</math> and substituting this into the second equation gives us an equation for <math>c</math>,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
3c(c-3c) &= -6\,,\\[5pt]
3c(c-3c) &= -6\,,\\[5pt]
-6c^2 &= -6\,,
-6c^2 &= -6\,,

Version vom 13:15, 10. Mär. 2009

A polynomial is said to have a triple root z=c if the equation contains the factor (z−c)3.

For our equation, this means that the left-hand side can be factorized as

z4−6z2+az+b=(z−c)3(z−d)

according to the factor theorem, where z=c is the triple root and z=d is the equation's fourth root (according to the fundamental theorem of algebra, a fourth-order equation always has four roots, taking into account multiplicity).

We will now try to determine a, b, c and d so that both sides in the factorization above agree.

If we expand the right-hand side above, we get

(z−c)3(z−d)=(z−c)2(z−c)(z−d)=(z2−2cz+c2)(z−c)(z−d)=(z3−3cz2+3c2z−c3)(z−d)=z4−(3c+d)z3+3c(c+d)z2−c2(c−3d)z+c3d

and this means that we must have

z4−6z2+az+b=z4−(3c+d)z3+3c(c+d)z2−c2(c−3d)z+c3d.

Because two polynomials are equal if an only if their coefficients are equal, this gives

3c+d3c(c+d)−c2(c−3d)c3d=0=−6=a=b.

From the first equation, we obtain d=−3c and substituting this into the second equation gives us an equation for c,

3c(c−3c)−6c2=−6=−6

i.e. c=−1 or c=1. The relation d=−3c gives that the corresponding values for d are d=3 and d=−3. The two last equations give us the corresponding values for a and b,


c=1 d=−3:abc=−1 d=3:ab=−12(1−3(−3))=8=13(−3)=−3=−(−1)2(−1−33)=10=(−1)33=−3.


Therefore, there are two different answers,

  • a=8 and b=−3 give the triple root z=1 and the single root z=−3,
  • a=10 and b=−3 give the triple root z=−1 and the single root z=3.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.4:5“