Question

The equation \(k-3x+x^{3} =0\). Find possible values of \(k\) so that the equation has at most \(2\) roots.

Answer 1

Well, how am I able to solve this since it's a cubic; I haven't done these in a while.

Anyone able to give me a hint?

Answer 2

Sure! Can we assume k is any constant # though?

Answer 3

Through trial and error?

Answer 4

All solutions are constants, aren't they?

Answer 5

Because if it's a variable this doesn't clean up neatly.

Answer 6

Trial-and-error is my preferable choice in the questions I can't solve :)

Answer 7

Well, I don't think trial and error is allowed. However, I do not know since I haven't done these in a very long time.

Answer 8

i think the grade of the x contained by k must be even

Answer 9

I am learning such questions, but they are quadratics.

Answer 10

If it was quadratics then discriminant would be used I think.

Answer 11

Yep

Answer 12

I am very rusty in these :(

Answer 13

no problem a polynome function has most 2 roots, when the highest polynom is even (e.g. 2,4,6,...)

Answer 14

which means you assume a solution..?

Answer 15

0 would be possible too or e.g. \[x^2 \]

Answer 16

but you do not know what is \(x\)

Answer 17

Explicit way to your answer
http://en.wikipedia.org/wiki/Cubic_function
Implicit way
we must think.....

Answer 18

sorry k= [x^2\] or k=[x^4\]

Answer 19

that is a rule

Answer 20

which means we guess an answer?

Answer 21

variable separable....

Answer 22

with an uneven polynome function you would get at most so much roots like the number of the highest polynom is

Answer 23

which basically means assumptions..
I don't you're allowed to assume in these question..

@lgbasallote: any ideas?

Answer 24

nawp. you guys are too smart for me

Answer 25

lol, this is elementary (:

Answer 26

\(k+x^3-3 x = 0\)
Subtract k from both sides:
\(-k = -3x+x^3\)
Multiply by (-1) to both sides:
\(k = 3x-x^3\)
Factor:
\(k = -x (x^2-3)\)
So
\(x = -k\)
or
\(x^2-3=k\)
\(x^2=k+3\)
\(x=\pm \sqrt{k+3}\)

So... That's 3 roots. But we can solve the roots implicitly by... http://en.wikipedia.org/wiki/Cubic_function#General_formula_of_roots , if you let it be in the form of ax^3+bx^2+cx+d=0, a\(\neq\)0

Answer 27

K=x^g for g is only an even figure and equal zero. Or am I absolutely wrong. you are starting to confuse

Answer 28

well, \(k\) is in terms of \(x\) then which is the roots?

Answer 29

only answers k=2 or -2

Answer 30

@Omniscience
have the answer?

Answer 31

Nope. :(
I will get the answer tomorrow though.

Answer 32

Arg I can't type fast enough...
If k = 2