next up previous
Next: Wielomiany stopnia 4 - Up: Wzory na pierwiastki wielomianów Previous: Wielomiany stopnia 2

Wielomiany stopnia 3 - wzory Cardano

Z uwagi na znaczną komplikację wzorów dla równania

ax3 + bx2 + cx + d = 0

w pełnej postaci:
x1 = $\displaystyle {\frac{{1}}{{3}}}$$\displaystyle \sqrt[3]{{\frac{9cba-27da^2-2b^3+3a\sqrt{-3\Delta}}{2a^3}}}$  
  - $\displaystyle \left(\vphantom{ \frac{3ca-b^2}{3a^2 \sqrt[3]{\frac{9cba-27da^2-2b^3+3a\sqrt{-3\Delta}}{2a^3}}}}\right.$$\displaystyle {\frac{{3ca-b^2}}{{3a^2 \sqrt[3]{\frac{9cba-27da^2-2b^3+3a\sqrt{-3\Delta}}{2a^3}}}}}$$\displaystyle \left.\vphantom{ \frac{3ca-b^2}{3a^2 \sqrt[3]{\frac{9cba-27da^2-2b^3+3a\sqrt{-3\Delta}}{2a^3}}}}\right)$ - $\displaystyle {\frac{{b}}{{3a}}}$  


x2 = $\displaystyle {\frac{{-1 + \sqrt{-3}}}{{6}}}$$\displaystyle \sqrt[3]{{\frac{9cba-27da^2-2b^3+3a\sqrt{-3\Delta}}{2a^3}}}$  
  + $\displaystyle {\frac{{1 + \sqrt{-3}}}{{6}}}$$\displaystyle \left(\vphantom{ \frac{3ca-b^2}{a^2 \sqrt[3]{\frac{9cba-27da^2-2b^3+3a\sqrt{-3\Delta}}{2a^3}}}}\right.$$\displaystyle {\frac{{3ca-b^2}}{{a^2 \sqrt[3]{\frac{9cba-27da^2-2b^3+3a\sqrt{-3\Delta}}{2a^3}}}}}$$\displaystyle \left.\vphantom{ \frac{3ca-b^2}{a^2 \sqrt[3]{\frac{9cba-27da^2-2b^3+3a\sqrt{-3\Delta}}{2a^3}}}}\right)$ - $\displaystyle {\frac{{b}}{{3a}}}$  


x3 = - $\displaystyle {\frac{{1 + \sqrt{-3}}}{{6}}}$$\displaystyle \sqrt[3]{{\frac{9cba-27da^2-2b^3+3a\sqrt{-3\Delta}}{2a^3}}}$  
  + $\displaystyle {\frac{{1 - \sqrt{-3}}}{{6}}}$$\displaystyle \left(\vphantom{ \frac{3ca-b^2}{a^2 \sqrt[3]{\frac{9cba-27da^2-2b^3+3a\sqrt{-3\Delta}}{2a^3}}}}\right.$$\displaystyle {\frac{{3ca-b^2}}{{a^2 \sqrt[3]{\frac{9cba-27da^2-2b^3+3a\sqrt{-3\Delta}}{2a^3}}}}}$$\displaystyle \left.\vphantom{ \frac{3ca-b^2}{a^2 \sqrt[3]{\frac{9cba-27da^2-2b^3+3a\sqrt{-3\Delta}}{2a^3}}}}\right)$ - $\displaystyle {\frac{{b}}{{3a}}}$  

gdzie

$\displaystyle \Delta$ = - 4c3a + c2b2 +18cbad - 27d2a2 -4db3

jest wyróżnikiem, równanie ax3 + bx2 + cx + d = 0 dzielimy przez a znajdując równoważne równanie x3 + (b/a)x2 + (c/a)x + (d /a) = 0, po czym podstawienie x = y - b/(3a)
$\displaystyle \left(\vphantom{y - \frac{b}{3a}}\right.$y - $\displaystyle {\frac{{b}}{{3a}}}$$\displaystyle \left.\vphantom{y - \frac{b}{3a}}\right)^{3}_{}$+$\displaystyle {\frac{{b}}{{a}}}$$\displaystyle \left(\vphantom{y - \frac{b}{3a}}\right.$y - $\displaystyle {\frac{{b}}{{3a}}}$$\displaystyle \left.\vphantom{y - \frac{b}{3a}}\right)^{2}_{}$+$\displaystyle {\frac{{c}}{{a}}}$$\displaystyle \left(\vphantom{y - \frac{b}{3a}}\right.$y - $\displaystyle {\frac{{b}}{{3a}}}$$\displaystyle \left.\vphantom{y - \frac{b}{3a}}\right)$+$\displaystyle {\frac{{d}}{{a}}}$=
  = y3 - $\displaystyle {\frac{{b^2}}{{3a^2}}}$y + $\displaystyle {\frac{{2b^3}}{{27a^3}}}$ + $\displaystyle {\frac{{c}}{{a}}}$y - $\displaystyle {\frac{{cb}}{{3a^2}}}$ + $\displaystyle {\frac{{d}}{{a}}}$ =  
  = y3 + $\displaystyle \left(\vphantom{ \frac{c}{a} - \frac{b^2}{3a^2}}\right.$$\displaystyle {\frac{{c}}{{a}}}$ - $\displaystyle {\frac{{b^2}}{{3a^2}}}$$\displaystyle \left.\vphantom{ \frac{c}{a} - \frac{b^2}{3a^2}}\right)$y - $\displaystyle {\frac{{cb}}{{3a^2}}}$ + $\displaystyle {\frac{{d}}{{a}}}$ + $\displaystyle {\frac{{2b^3}}{{27a^3}}}$  

sprowadza równanie do postaci y3 + py + q = 0, i znajdujemy rozwiązania z wzorów Cardano:

y1 = $\displaystyle \left(\vphantom{ \sqrt[3]{\frac{-9q + \sqrt{-3\Delta}}{18}} }\right.$$\displaystyle \sqrt[3]{{\frac{-9q + \sqrt{-3\Delta}}{18}}}$$\displaystyle \left.\vphantom{ \sqrt[3]{\frac{-9q + \sqrt{-3\Delta}}{18}} }\right)$ - $\displaystyle \left(\vphantom{ \frac{p}{3 \frac{-9q + \sqrt{-3\Delta}}{18}}}\right.$$\displaystyle {\frac{{p}}{{3 \frac{-9q + \sqrt{-3\Delta}}{18}}}}$$\displaystyle \left.\vphantom{ \frac{p}{3 \frac{-9q + \sqrt{-3\Delta}}{18}}}\right)$,

y2 = $\displaystyle {\frac{{-1 + \sqrt{-3}}}{{2}}}$$\displaystyle \left(\vphantom{ \sqrt[3]{\frac{-9q + \sqrt{-3\Delta}}{18}} }\right.$$\displaystyle \sqrt[3]{{\frac{-9q + \sqrt{-3\Delta}}{18}}}$$\displaystyle \left.\vphantom{ \sqrt[3]{\frac{-9q + \sqrt{-3\Delta}}{18}} }\right)$ + $\displaystyle {\frac{{1 + \sqrt{-3}}}{{2}}}$$\displaystyle \left(\vphantom{ \frac{p}{3 \frac{-9q + \sqrt{-3\Delta}}{18}}}\right.$$\displaystyle {\frac{{p}}{{3 \frac{-9q + \sqrt{-3\Delta}}{18}}}}$$\displaystyle \left.\vphantom{ \frac{p}{3 \frac{-9q + \sqrt{-3\Delta}}{18}}}\right)$,

y3 = $\displaystyle {\frac{{-1 - \sqrt{-3}}}{{2}}}$$\displaystyle \left(\vphantom{ \sqrt[3]{\frac{-9q + \sqrt{-3\Delta}}{18}} }\right.$$\displaystyle \sqrt[3]{{\frac{-9q + \sqrt{-3\Delta}}{18}}}$$\displaystyle \left.\vphantom{ \sqrt[3]{\frac{-9q + \sqrt{-3\Delta}}{18}} }\right)$ + $\displaystyle {\frac{{1 - \sqrt{-3}}}{{2}}}$$\displaystyle \left(\vphantom{ \frac{p}{3 \frac{-9q + \sqrt{-3\Delta}}{18}}}\right.$$\displaystyle {\frac{{p}}{{3 \frac{-9q + \sqrt{-3\Delta}}{18}}}}$$\displaystyle \left.\vphantom{ \frac{p}{3 \frac{-9q + \sqrt{-3\Delta}}{18}}}\right)$,

gdzie

D = - 4p3 -27q2

jest wyróżnikiem.

Jeśli nie mamy ochoty uczyć się tych wzorów, to - metodą Thomasa Harriota - w równaniu y3 + py + q = 0 podstawiamy y = z - p/(3z):

$\displaystyle \left(\vphantom{ z - \frac{p}{3z} }\right.$z - $\displaystyle {\frac{{p}}{{3z}}}$$\displaystyle \left.\vphantom{ z - \frac{p}{3z} }\right)^{3}_{}$ + p$\displaystyle \left(\vphantom{ z - \frac{p}{3z} }\right.$z - $\displaystyle {\frac{{p}}{{3z}}}$$\displaystyle \left.\vphantom{ z - \frac{p}{3z} }\right)$ + q = z3 - $\displaystyle {\frac{{p^3}}{{27z^3}}}$ + q

i t = z3 znajdujemy, rozwiązując równanie

t2 + qt - $\displaystyle {\frac{{p^3}}{{27}}}$ = 0


next up previous
Next: Wielomiany stopnia 4 - Up: Wzory na pierwiastki wielomianów Previous: Wielomiany stopnia 2
Pawel Gladki 2006-01-30