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I don't know what the original cubic was. I got interested in this by viewing this youtube video:
https://www.youtube.com/watch?v=_qvp9a1x2UM
For the original cubic, it must have been x^3 + ax + b
If my reverse engineering is correct, then b=-10 and a = -3*cuberoot(40)
The links you posted look very helpful. Thanks.
Yes. He set t=u+v and expanded the expression (u+v)^3 + p(u+v) + q = u^3 + 3uv^2 + 3(u^2)v + v^3 +pu + pv + q
Simplifying, this becomes u^3 + v^3 + q + (3uv)v + (3uv)u +p(u+v) = u^3 + v^3 + q + (3uv+p)(u+v)
Then he set 3uv+p = 0 and set r=u^3, s=v^3, to get the two equations:
(1) rs = (uv)^3 = (-p/3)^3 = -p^3/27
and
(2) r+s = u^3 + v^3 = -q
He then substituted r = -(q+s) into equation (1) to get
(3) -(q+s)s = -p^3/27
which is a quadratic in s (q and p are the known coefficients from the original cubic, s is the unknown variable).
If this quadratic has real roots, then the solution to the cubic is straightforward: solve for r and s, and set t = cuberoot(r)+cuberoot(s).
But it is my understanding that Cardano found an equation for which r = 5 + sqrt(-15), s = 5 - sqrt(-15), and still found real roots of the cubic function, even though he described it as "mental torture" to work with what he called an "imaginary" number.
Is this true? If so, how did he do it?
I get stuck at the line "At this point, Cardano, who did not know complex numbers, supposed that the roots of this equation were real, that is that q^2/4 + p^3/27 >0 "
My understanding is that Cardano came across the situation where q^2/4 + p^3/27 = -15, and this was the first instance ever of an imaginary number being used in a calculation.
Did Cardano, in fact, solve a cubic equation by treating sqrt(-15) as if it were a usable number? If so, how did he solve it?
Given a cubic polynomial of the form x^3+ax+b=0, I understand that Cardano solved it by substituting x=(p+q) and setting 3pq+a=0, or pq=-a/3. He then set u=p^3 and v=q^3, converting it into a quadratic equation in v.
u + v = -b
uv = -(a^3/27)
uv = -(b+v)v = -(a^3/27)
v^2 +bv = a^3/27
As long as this quadratic equation had real roots, he could find x=cuberoot(v) - a/3*(cuberoot(v)). (Yes, I know that Tartaglia is believed to be the one who came up with this method.)
Cardano became the first mathematician to make use of what he called an "imaginary number", the sqrt(-15), when he encountered a quadratic equation with roots v=5+sqrt(-15), and u=5-sqrt(-15), giving x=cuberoot(v) - a/(3*cuberoot(v))
This much, I understand.
What I can't figure out is how Cardano derived a real value for x from the difference of two cube roots of complex numbers.
Can anyone explain to me how Cardano went from v=5+sqrt(-15) to a real value for x?
Thanks.
For #2, forget about the line being inside a square. Look at it as the straight line y=.75x, and ask: after (0,0), what is the first point on the line where x,y are both integers.
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