# depressed cubics

November 1, 2010

Aim: Solve cubics of form
$x^2=fx+g$

Do Now: Solve by "guess and check"
1)
$x^3=12x+65\\ (5)^3=12(5)+65\\ x=5$

2)
$x^3=6x+9\\ (3)^3=6(3)+9\\ x=3$

3)
$x^3=6x+40\\ (4)^3=6(4)+40\\ x=4$

4) Cubics must have at least 1 root, root(s) for this equation are irrational (none of the potential roots work)

$x^3=15x+70\\\\ \sqrt[3]{(70/2)+\sqrt{(70/2)^2-(15/3)^3}}+\sqrt[3]{(70/2)-\sqrt{(70/2)^2-(15/3)^3}} ==>\\\\ \sqrt[3]{35+\sqrt1100}+\sqrt[3]{35-\sqrt1100}$

Refer to Solving the Depressed Cubic

In contrast to the quadratic formula,

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

there is also a cubic formula (Cardano's Formula) developed by Cardano and his student Ferrari.

$x=\sqrt[3]{(\dfrac{-b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a})+\sqrt{(\dfrac{-b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a})^2+(\dfrac{c}{3a}-\dfrac{b^2}{9a^2})^3}}\\ +\sqrt[3]{(\dfrac{-b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a})-\sqrt{(\dfrac{-b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a})^2+(\dfrac{c}{3a}-\dfrac{b^2}{9a^2})^3}}-\dfrac{b}{3a}}$

Solving the Depressed Cubic

$Given\;the\;form:\;x^3=fx+g\\\\ Solution:\sqrt[3]{(g/2)+\sqrt{(g/2)^2-(f/3)^3}}+\sqrt[3]{(g/2)-\sqrt{(g/2)^2-(f/3)^3}}$

This formula only works if (f/3)^3 is less than (g/2)^2

Explanation in words : The solution will always take the form above. The constant under the radical is one half the g term in the equation and the other part under the radical is obtained by subtracting (f/3)^3 from (g/2)^2. The second radical is the same as the first but with the opposite sign under the radical.

Euler's Method: (when given 3ab=f and a^3+b+3=g)

$x^3=6x+40$

$x=a+b\\\\ x^3=(a+b)^3=a^3+3a^2b+3ab^2+b^3\\\\ =3ab(a+b)+a^3+b^3\\\\ =(3ab)x+(a^3+b^3)$

We know that : 3ab = 6 and a^3+b^3=40
(b = 6/3a => b = 2/a)

$a^3+(\dfrac{2}{a})^3=40\\\\ =a^3+\dfrac{8}{a^3}=40\\\\ =a^6+8=40a^3\\\\ =a^6-40a^3+8=0$

Using the quadratic formula, we end up with...

$a^3=\dfrac{40+\sqrt1568}{2}\\\\ a^3=20+\sqrt392\\\\ a=\sqrt[3]{20+\sqrt392}}$

To solve for b...

$a^3+b^3=40\\\\ (20+\sqrt392)+b^3=40\\\\ b=20-\sqrt392\\\\$

Check :

$a+b=40\\\\ (20+\sqrt392)+(20-\sqrt392)=40\\\\ 40=40$

Gary Wu
Period 3