If x^3+ax+b=0, & a,b belongs to real numbers, b is not equal to zero.Then, why a

Question

If x^3+ax+b=0, & a,b belongs to real numbers, b is not equal to zero.Then, why a<0,since it is the a statement which holds true.And all the roots of this equation are real.

blockcolder · Answer

So your question is...?

maheshmeghwal9 · Answer

my ques is why a<0.under what coditions?

maheshmeghwal9 · Answer

sorry,under what conditions?

maheshmeghwal9 · Answer

this is a question of theory of equations.

blockcolder · Answer

Do you have other assumptions on the cubic equation, like, how many solutions does it have, or other stuff like that?

maheshmeghwal9 · Answer

no I don't know about the solutions but I know that this is a thoughtful question.

maheshmeghwal9 · Answer

this is only verify and prove question.

anonymous · Answer

The statement is not complete.
For example: we say the standard form of quadratic equation is 
$$ax^2+bx+c=0, where a, b, c \in R $$ and a, b not both zero

maheshmeghwal9 · Answer

but question is is true because my sir has solved it and i have lost that paper on which sir solved that.

freckles · Answer

I don't understand the question. You want us to solve for x? Is that what your asking?

anonymous · Answer

http://www.sosmath.com/algebra/factor/fac11/fac11.html

maheshmeghwal9 · Answer

what is this?

maheshmeghwal9 · Answer

one more thing about this quetion

maheshmeghwal9 · Answer

all the roots of this cubic equations are real

freckles · Answer

Oh ok now the question makes more sense. lol/

freckles · Answer

Omg I'm sorry. My page got killed when I was typing all of that. :(

maheshmeghwal9 · Answer

np!

freckles · Answer

@foolformath what do you think of this problem?
For some reason I was thinking about finding f'
And I found critical numbers pm sqrt(-a/3)

I found it is increasing on (-inf,-sqrt(-a/3)) and (sqrt(-a/3),inf)

and decreasing on (-sqrt(-a/3),sqrt(-a/3))


I can't decide what to do from this or if i can do anything with what i found....

maheshmeghwal9 · Answer

ok np. :(

freckles · Answer

i'm still thinking 
:(

freckles · Answer

@KingGeorge any thoughts?

kinggeorge · Answer

Sorry for asking this, but just to clarify, you're asking for conditions under which a will be less than 0 correct?

freckles · Answer

What do you think about taking the contrapositive?

kinggeorge · Answer

If that's the question, looking at the derivative is a good idea. $$f(x)=x^3+ax+b$$$$f'(x)=3x^2+a$$If $a$ is negative, then we would have 2 critical points that translate to local maxima and local minima. If a is positive, we would have 0 critical points.

kinggeorge · Answer

If we drew the graphs, $a$ negative would look similar to this:|dw:1336193429080:dw|And $a$ positive would look someting like|dw:1336193456467:dw|