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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.
 one year ago
 one year ago
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.
 one year ago
 one year ago

This Question is Closed

blockcolderBest ResponseYou've already chosen the best response.0
So your question is...?
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
my ques is why a<0.under what coditions?
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
sorry,under what conditions?
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
this is a question of theory of equations.
 one year ago

blockcolderBest ResponseYou've already chosen the best response.0
Do you have other assumptions on the cubic equation, like, how many solutions does it have, or other stuff like that?
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
no I don't know about the solutions but I know that this is a thoughtful question.
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
this is only verify and prove question.
 one year ago

veramathBest ResponseYou've already chosen the best response.0
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
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
but question is is true because my sir has solved it and i have lost that paper on which sir solved that.
 one year ago

frecklesBest ResponseYou've already chosen the best response.1
I don't understand the question. You want us to solve for x? Is that what your asking?
 one year ago

cinarBest ResponseYou've already chosen the best response.0
http://www.sosmath.com/algebra/factor/fac11/fac11.html
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
one more thing about this quetion
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
all the roots of this cubic equations are real
 one year ago

frecklesBest ResponseYou've already chosen the best response.1
Oh ok now the question makes more sense. lol/
 one year ago

frecklesBest ResponseYou've already chosen the best response.1
Omg I'm sorry. My page got killed when I was typing all of that. :(
 one year ago

frecklesBest ResponseYou've already chosen the best response.1
@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....
 one year ago

frecklesBest ResponseYou've already chosen the best response.1
i'm still thinking :(
 one year ago

frecklesBest ResponseYou've already chosen the best response.1
@KingGeorge any thoughts?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.2
Sorry for asking this, but just to clarify, you're asking for conditions under which a will be less than 0 correct?
 one year ago

frecklesBest ResponseYou've already chosen the best response.1
What do you think about taking the contrapositive?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.2
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.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.2
If we drew the graphs, \(a\) negative would look similar to this:dw:1336193429080:dwAnd \(a\) positive would look someting likedw:1336193456467:dw
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.2
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.
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
you both are good helper.
 one year ago
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