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# Discriminant watch

1. How do I derive the dicriminant of a cubic function??
i.e ,
2. What's your definition of "discriminant"?
3. 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
4. (Original post by generalebriety)
ermm, I dont know what the definition is .. all I know is it can be used to find the nature of stationary points ..
5. (Original post by rbnphlp)
ermm, I dont know what the definition is .. all I know is it can be used to find the nature of stationary points ..
Well, that'll be why you can't derive it...

What level are you at? GCSE / AS / A2?
6. (Original post by 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

..

I just finished my Alevels
7. B^2-4ac is the discriminant.
8. (Original post by learner_dancer)
B^2-4ac is the discriminant.
not for a cubic function it isnt.
9. (Original post by rbnphlp)
not for a cubic function it isnt.
Nope.
Sigh.
10. (Original post by 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.
11. (Original post by rbnphlp)
ok let me rephrase my question How would I solve a cubic finction

..

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.)
12. wiki fail you mean.
13. (Original post by 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

14. (Original post by rbnphlp)
I thought of that , but it doesnt work for the equation i wanted to solve
What?
15. (Original post by 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/...+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.
16. (Original post by rbnphlp)
I thought of that , but it doesnt work for the equation i wanted to solve

17. (Original post by Simplicity)
Wolfram alpha gives this
http://www76.wolframalpha.com/input/...+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.
No I i didnt make it up .. I came Across it on " the history of mathematics"
18. Another way to solve it is to introduce a linear shift y = x-p to 'remove' the x^2 term, to then get you an equation of the form . Then introduce another substitution of the form z=qy to transform the above equation into something of the form , and then note that a solution to this is given by either cos (arcos k)/3 , cosh (arcosh k)/3, or sinh (arsinh k)/3 depending on whether you have a +, - and whether k is greater or smaller than 1.

i'm not too sure of the above, but i think the general gist of it is right.

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