Hi,
I created the quartic equation \(\displaystyle 36x^4-72x^3-391x^2-123x+270=0\)
by multiplying out \(\displaystyle (3x-2)(2x+3)(3x+5)(2x-9)=0\) so that I know the roots are \(\displaystyle x=\dfrac{2}{3}\ \ \ x=-\dfrac{3}{2}\ \ \ x=-\dfrac{5}{3}\ \ \ and\ \ \ x=\dfrac{9}{2}\)
I then tried following Ferrari's method from http://en.wikipedia.org/wiki/Quartic_formula#Summary_of_Ferrari.27s_method to work from the quartic to get the roots. I'm apparently doing something wrong as I can get nowhere near a solution
I have \(\displaystyle A=36\ \ \ B=-72\ \ \ C=-391\ \ \ D=-123\ \ \ E=270\)
plugging these in I get . . .
\(\displaystyle \alpha=\dfrac{C}{A}-\dfrac{3B^2}{8A^2}=-\dfrac{445}{36}\)
\(\displaystyle \beta=\dfrac{B^3}{8A^3}-\dfrac{BC}{2A^2}+\dfrac{D}{A}=-\dfrac{275}{18}\)
\(\displaystyle \gamma=\dfrac{CB^2}{16A^3}-\dfrac{3B^4}{256A^4}-\dfrac{BD}{4A^2}+\dfrac{E}{A}=\dfrac{26}{9}\)
\(\displaystyle P=-\dfrac{\alpha^2}{12}-\gamma=-\dfrac{242953}{15552}\)
\(\displaystyle Q=\dfrac{\alpha\gamma}{3}-\dfrac{\beta^2}{8}-\dfrac{\alpha^3}{108}=-\dfrac{118872755}{5038848}\)
There's nothing in the above that can cause any problems but the next stage . . .
\(\displaystyle R=-\dfrac{Q}{2}\pm\sqrt{\dfrac{Q^2}{4}+\dfrac{P^3}{27}}\)
this produces a complex R (i.e. the square-root part is negative).
continuing on with the next three steps does not remove this complex element and so doesn't give any of the initial roots.
Am I doing this incorrectly or have I misunderstood the method ?
Many thanks.
I created the quartic equation \(\displaystyle 36x^4-72x^3-391x^2-123x+270=0\)
by multiplying out \(\displaystyle (3x-2)(2x+3)(3x+5)(2x-9)=0\) so that I know the roots are \(\displaystyle x=\dfrac{2}{3}\ \ \ x=-\dfrac{3}{2}\ \ \ x=-\dfrac{5}{3}\ \ \ and\ \ \ x=\dfrac{9}{2}\)
I then tried following Ferrari's method from http://en.wikipedia.org/wiki/Quartic_formula#Summary_of_Ferrari.27s_method to work from the quartic to get the roots. I'm apparently doing something wrong as I can get nowhere near a solution
I have \(\displaystyle A=36\ \ \ B=-72\ \ \ C=-391\ \ \ D=-123\ \ \ E=270\)
plugging these in I get . . .
\(\displaystyle \alpha=\dfrac{C}{A}-\dfrac{3B^2}{8A^2}=-\dfrac{445}{36}\)
\(\displaystyle \beta=\dfrac{B^3}{8A^3}-\dfrac{BC}{2A^2}+\dfrac{D}{A}=-\dfrac{275}{18}\)
\(\displaystyle \gamma=\dfrac{CB^2}{16A^3}-\dfrac{3B^4}{256A^4}-\dfrac{BD}{4A^2}+\dfrac{E}{A}=\dfrac{26}{9}\)
\(\displaystyle P=-\dfrac{\alpha^2}{12}-\gamma=-\dfrac{242953}{15552}\)
\(\displaystyle Q=\dfrac{\alpha\gamma}{3}-\dfrac{\beta^2}{8}-\dfrac{\alpha^3}{108}=-\dfrac{118872755}{5038848}\)
There's nothing in the above that can cause any problems but the next stage . . .
\(\displaystyle R=-\dfrac{Q}{2}\pm\sqrt{\dfrac{Q^2}{4}+\dfrac{P^3}{27}}\)
this produces a complex R (i.e. the square-root part is negative).
continuing on with the next three steps does not remove this complex element and so doesn't give any of the initial roots.
Am I doing this incorrectly or have I misunderstood the method ?
Many thanks.
