| Author |
Message |
Steve Hewson
Frequent poster
Post Number: 143
| | Posted on Monday, 15 November, 2010 - 10:15 am: | |
If you've got any comments on this problem or it prompts you into other investigations, please post them here.
|
azerbajdzan
Veteran poster
Post Number: 531
| | Posted on Monday, 22 November, 2010 - 01:10 pm: | |
Although it wasn't purpose of this challenge to find exact formulas, just in case someone is interested...
[display]2 x^3+34 x^2+567 x+8901=0[/display]
[display]x_1=-\frac{17}{3}+\sqrt[3]{\frac{1}{2} \sqrt{\frac{67572913}{6}}-\frac{43307}{27}}-\sqrt[3]{\frac{1}{2} \sqrt{\frac{67572913}{6}}+\frac{43307}{27}}[/display]
[display]x_2=-\frac{17}{3}-\left(1+i \sqrt{3}\right) \sqrt[3]{\frac{1}{16} \sqrt{\frac{67572913}{6}}-\frac{43307}{216}}+\left(1-i \sqrt{3}\right) \sqrt[3]{\frac{1}{16} \sqrt{\frac{67572913}{6}}+\frac{43307}{216}}[/display]
[display]x_3=-\frac{17}{3}-\left(1-i \sqrt{3}\right) \sqrt[3]{\frac{1}{16} \sqrt{\frac{67572913}{6}}-\frac{43307}{216}}+\left(1+i \sqrt{3}\right) \sqrt[3]{\frac{1}{16} \sqrt{\frac{67572913}{6}}+\frac{43307}{216}}[/display]
Formulas are adapted in such a form, that numbers under all roots are positive, so you can use common calculator to compute the result without problems of imaginary parts of numbers |
azerbajdzan
Veteran poster
Post Number: 532
| | Posted on Monday, 22 November, 2010 - 01:54 pm: | |
...or a better form for computation, because it uses only one shape of roots
[display]x_1=-\frac{17}{3}+\sqrt[3]{\frac{1}{2} \sqrt{\frac{67572913}{6}}-\frac{43307}{27}}-\frac{1123}{18 \sqrt[3]{\frac{1}{2} \sqrt{\frac{67572913}{6}}-\frac{43307}{27}}}[/display]
[display]x_1=-\frac{17}{3}+r-\frac{1123}{18 r}[/display]
[display]r=\sqrt[3]{\frac{1}{2} \sqrt{\frac{67572913}{6}}-\frac{43307}{27}}[/display] |
Steve Hewson
Frequent poster
Post Number: 147
| | Posted on Monday, 22 November, 2010 - 02:03 pm: | |
@azerbajdzan I'm glad that you mentioned the formula for the cubics and had the resiliance to encode in Latex!
One can also consider the discriminant of a cubic [inline]ax^3+bx^2+cx+d[/inline]
[display]b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd[/display] to determine the number of real solutions.
|
azerbajdzan
Veteran poster
Post Number: 533
| | Posted on Monday, 22 November, 2010 - 02:44 pm: | |
In fact they all can be expressed using the r from above post:
[display]2 x^3+34 x^2+567 x+8901=0[/display]
[display]r=\sqrt[3]{\frac{1}{2} \sqrt{\frac{67572913}{6}}-\frac{43307}{27}}[/display]
[display]x_1=-\frac{17}{3}-\left(\frac{1123}{18 r}-r\right)[/display]
[display]x_2=-\frac{17}{3}+\frac{1}{2} \left(\frac{1123}{18 r}-r\right)-i \frac{\sqrt{3}}{2} \left(\frac{1123}{18 r}+r\right)[/display]
[display]x_3=-\frac{17}{3}+\frac{1}{2} \left(\frac{1123}{18 r}-r\right)+i \frac{\sqrt{3}}{2} \left(\frac{1123}{18 r}+r\right)[/display]
discriminant is -4865249736 if I did not make a mistake... I think it can be negative only when there are two complex and one real solution in cubic...but not sure |