Discriminant

one must think of a cubic
with graphic properties
and shape and turning points
most appropriately
so I would recommend you derive
and solve equal to nought
and with these station'ry values
seek the solution that was sought

Reply 3

OP

generalebriety

What's your definition of "discriminant"?

ermm, I dont know what the definition is .. all I know is it can be used to find the nature of stationary points :o:

..

Reply 4

generalebriety

16

rbnphlp

ermm, I dont know what the definition is .. all I know is it can be used to find the nature of stationary points :o:

..

Well, that'll be why you can't derive it...

What level are you at? GCSE / AS / A2?

Reply 5

OP

generalebriety

Well, that'll be why you can't derive it...

What level are you at? GCSE / AS / A2?

ok let me rephrase my question How would I solve a cubic finction

ax^3+bx^2+cx+d

..

I just finished my Alevels

Reply 6

learner_dancer

12

B^2-4ac is the discriminant.

Reply 7

OP

learner_dancer

B^2-4ac is the discriminant.

not for a cubic function it isnt. :p:

Reply 8

learner_dancer

12

rbnphlp

not for a cubic function it isnt. :p:

Nope.
Sigh.

Reply 9

Simplicity

rbnphlp

ok let me rephrase my question How would I solve a cubic finction

..

I just finished my Alevels

Newton raphson?

There is a general formula but its very long and impossible to memorise. Your best bet would be to find a factor then do polynomial divison.

Reply 10

generalebriety

16

rbnphlp

ok let me rephrase my question How would I solve a cubic finction

ax^3+bx^2+cx+d

..

I just finished my Alevels

That's a completely different question. See here.

For your original question (though be warned that this will require you to do a lot of algebra): take two polynomials P and Q. Consider the product of all (x - y), where P(x) = 0 and Q(y) = 0, i.e. x is a root of P and y is a root of Q. Obviously this product is 0 if any one of its factors is 0, i.e. if any one root of P happens to also be one root of Q, and is non-zero otherwise. Now suppose we want to know whether P has a repeated root, say r. Well, if it does, then P is 0 at r, but P also turns at r, so P' (the derivative of P) is 0 at r. So let Q = P' in the above quantity. But of course we don't know the roots yet - so what can we do? Well, suppose P has roots r, s, t; then by writing P = ax^3 + bx^2 + cx + d = a(x-r)(x-s)(x-t) we can find that -d/a = rst, c/a = rs+rt+st, -b/a = r+s+t; from there on it's all just horrible algebra. (If you're interested in a less hurried explanation of this, search Wikipedia or something for the "resultant", a related quantity.)

wiki fail you mean.

OP

Simplicity

Newton raphson?

There is a general formula but its very long and impossible to memorise. Your best bet would be to find a factor then do polynomial divison.

I thought of that , but it doesnt work for the equation i wanted to solve

x^3+x=6

Reply 13

Simplicity

rbnphlp

I thought of that , but it doesnt work for the equation i wanted to solve

What?

Reply 14

Simplicity

rbnphlp

I thought of that , but it doesnt work for the equation i wanted to solve

Wolfram alpha gives this
http://www76.wolframalpha.com/input/?i=root+x%5E3%2Bx%3D6

P.S. Where did you get the problem. As I'm pretty sure no one would ask you that in a textbook. So you must have made that up.

rbnphlp

I thought of that , but it doesnt work for the equation i wanted to solve

x^3+x=6

google depressed cubic.

Reply 16

OP