- #1
DrummingAtom
- 657
- 2
I got this out of An Imaginary Tale: The Story of Sqrt(-1). In section 1.5 of the book, the author explains that Bombelli took x3 = 15x + 4 and found the real solutions: 4, -2±sqrt(3). But if you plug the equation into the Cardan formula you get imaginaries. http://en.wikipedia.org/wiki/Cardan_formula#Cardano.27s_method
The author shows that if a and b are some yet to be determined real numbers where:
\sqrt[3]{2+\sqrt{-121}} = a+b\sqrt{-1}
\sqrt[3]{2-\sqrt{-121}} = a-b\sqrt{-1}
Then he takes the first equation and cubes both sides, does a bunch of Algebra and gets:
2+\sqrt{-121} = a(a^2-3b^2)+b(3a^2-b^2)\sqrt{-1}
And says if this is equal to the complex number, 2+\sqrt{-121} then the real and imaginary parts must be separately equal. Then he splits terms into:
a(a^2-3b^2) = 2
b(3a^2-b^2)\sqrt{-1}=11
To find that a = 2 and b = 1, then says "With these results Bombelli showed that the Cardan solution is 4 and this is correct."
The very last part is where I don't understand how all that complex stuff arrives back at 4. Even though, through simple Algebra with the very first equation with have real solutions.. Any help would be awesome. Thanks.
The author shows that if a and b are some yet to be determined real numbers where:
\sqrt[3]{2+\sqrt{-121}} = a+b\sqrt{-1}
\sqrt[3]{2-\sqrt{-121}} = a-b\sqrt{-1}
Then he takes the first equation and cubes both sides, does a bunch of Algebra and gets:
2+\sqrt{-121} = a(a^2-3b^2)+b(3a^2-b^2)\sqrt{-1}
And says if this is equal to the complex number, 2+\sqrt{-121} then the real and imaginary parts must be separately equal. Then he splits terms into:
a(a^2-3b^2) = 2
b(3a^2-b^2)\sqrt{-1}=11
To find that a = 2 and b = 1, then says "With these results Bombelli showed that the Cardan solution is 4 and this is correct."
The very last part is where I don't understand how all that complex stuff arrives back at 4. Even though, through simple Algebra with the very first equation with have real solutions.. Any help would be awesome. Thanks.