Weekly Challenge 10: Solve me Log Out | Topics | Search Moderators | Register | Edit Profile

 Ask NRICH » Onwards and Upwards » Weekly Challenge 10: Solve me « Previous Next »

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...

$$2 x^3+34 x^2+567 x+8901=0$$
$$x_1=-\frac{17}{3}+\sqrt[3]{\frac{1}{2} \sqrt{\frac{67572913}{6}}-\frac{43307}{27}}-\sqrt[3]{\frac{1}{2} \sqrt{\frac{67572913}{6}}+\frac{43307}{27}}$$
$$x_2=-\frac{17}{3}-\left(1+i \sqrt{3}\right) \sqrt[3]{\frac{1}{16} \sqrt{\frac{67572913}{6}}-\frac{43307}{216}}+\left(1-i \sqrt{3}\right) \sqrt[3]{\frac{1}{16} \sqrt{\frac{67572913}{6}}+\frac{43307}{216}}$$
$$x_3=-\frac{17}{3}-\left(1-i \sqrt{3}\right) \sqrt[3]{\frac{1}{16} \sqrt{\frac{67572913}{6}}-\frac{43307}{216}}+\left(1+i \sqrt{3}\right) \sqrt[3]{\frac{1}{16} \sqrt{\frac{67572913}{6}}+\frac{43307}{216}}$$

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

$$x_1=-\frac{17}{3}+\sqrt[3]{\frac{1}{2} \sqrt{\frac{67572913}{6}}-\frac{43307}{27}}-\frac{1123}{18 \sqrt[3]{\frac{1}{2} \sqrt{\frac{67572913}{6}}-\frac{43307}{27}}}$$

$$x_1=-\frac{17}{3}+r-\frac{1123}{18 r}$$
$$r=\sqrt[3]{\frac{1}{2} \sqrt{\frac{67572913}{6}}-\frac{43307}{27}}$$
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 $ax^3+bx^2+cx+d$
$$b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$$ 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:
$$2 x^3+34 x^2+567 x+8901=0$$
$$r=\sqrt[3]{\frac{1}{2} \sqrt{\frac{67572913}{6}}-\frac{43307}{27}}$$
$$x_1=-\frac{17}{3}-\left(\frac{1123}{18 r}-r\right)$$
$$x_2=-\frac{17}{3}+\frac{1}{2} \left(\frac{1123}{18 r}-r\right)-i \frac{\sqrt{3}}{2} \left(\frac{1123}{18 r}+r\right)$$
$$x_3=-\frac{17}{3}+\frac{1}{2} \left(\frac{1123}{18 r}-r\right)+i \frac{\sqrt{3}}{2} \left(\frac{1123}{18 r}+r\right)$$

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