- #1
Char. Limit
Gold Member
- 1,222
- 22
The proof of the quadratic formula was so simple, I moved to the proof of the cubic formula with supreme confidence. And found myself awash in as and cs and cubic roots.
Can you turn this equation into a cubic?
[tex]x=-\frac{b}{3a}[/tex]
[tex]-\frac{1}{3a}\sqrt[3]{\frac{1}{2}(2b^3-9abc+27(a^2)d+\sqrt{(2b^3-9abc+27(a^2)d)^2-4(b^2-3ac)^3}}[/tex]
[tex]-\frac{1}{3a}\sqrt[3]{\frac{1}{2}(2b^3-9abc+27(a^2)d-\sqrt{(2b^3-9abc+27(a^2)d)^2-4(b^2-3ac)^3}}[/tex]
Can you turn this equation into a cubic?
[tex]x=-\frac{b}{3a}[/tex]
[tex]-\frac{1}{3a}\sqrt[3]{\frac{1}{2}(2b^3-9abc+27(a^2)d+\sqrt{(2b^3-9abc+27(a^2)d)^2-4(b^2-3ac)^3}}[/tex]
[tex]-\frac{1}{3a}\sqrt[3]{\frac{1}{2}(2b^3-9abc+27(a^2)d-\sqrt{(2b^3-9abc+27(a^2)d)^2-4(b^2-3ac)^3}}[/tex]