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.

- maheshmeghwal9

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- blockcolder

So your question is...?

- maheshmeghwal9

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

- maheshmeghwal9

sorry,under what conditions?

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## More answers

- maheshmeghwal9

this is a question of theory of equations.

- blockcolder

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

- maheshmeghwal9

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

- maheshmeghwal9

this is only verify and prove question.

- anonymous

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

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

- freckles

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

- anonymous

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

- maheshmeghwal9

what is this?

- maheshmeghwal9

one more thing about this quetion

- maheshmeghwal9

all the roots of this cubic equations are real

- freckles

Oh ok now the question makes more sense. lol/

- freckles

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

- maheshmeghwal9

np!

- freckles

@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

ok np. :(

- freckles

i'm still thinking
:(

- freckles

@KingGeorge any thoughts?

- KingGeorge

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

- freckles

What do you think about taking the contrapositive?

- KingGeorge

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

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|

- KingGeorge

This means that if \(a\) were positive, then the function would always be increasing. Hence, we would only have one real root. But we're required to have all real roots. Contradiction!
Therefore, a must be less than 0.

- maheshmeghwal9

thanx!

- KingGeorge

You're welcome.

- freckles

yes thanks george :)

- maheshmeghwal9

you both are good helper.

- freckles

george is bester!

- KingGeorge

Thank you guys :)

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