The Student Room Group
Reply 1
Solve
u-v=d
and
uv=(c/3)^3
simultaneously.

(For a given equation you will know c and d.)
Reply 2
rnd
Solve
u-v=d
and
uv=(c/3)^3
simultaneously.

(For a given equation you will know c and d.)

My question is how do I deirive [ u-v=d , and uv = (c/3)^3,
Unparseable latex formula:

x=\sqrt[3]{u}-\sqrt[3]{v}}

}.


given only that ax3+cx=dax^3+cx=d
soory if my question isnt specific
Reply 3
Are you sure you want to derive those equations?

If you just want to solve them then you have u=d+v, sub that in uv=(c/3)^3 rarrange to get a quadratic equation and solve it to get v in terms of c and d.

But how can you not be using a?
Reply 4
yes, I want to know how they found the equations(I think the bok assumes its obvious :rolleyes: .I know how to find x using u and v.
Reply 5
rbnphlp
yes, I want to know how they found the equations(I think the bok assumes its obvious :rolleyes: .I know how to find x using u and v.


OK. I'm giving up on this one then. Hope someone else helps you out. :smile:

My point about a still stands though. How can x depend on c and d but not a? It can't. What's the book by the way?
Reply 6
It looks like the standard cubic solution equation in the special case where b=0 and a=1. There's a derivation of the solution on this page:

http://en.wikipedia.org/wiki/Cubic_equation#Roots_of_a_cubic_function
Reply 7
rnd
OK. I'm giving up on this one then. Hope someone else helps you out. :smile:

My point about a still stands though. How can x depend on c and d but not a? It can't. What's the book by the way?

History of mathematics :smile: quite intresting

Latest