Here's the question you clicked on:
maheshmeghwal9
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.
So your question is...?
my ques is why a<0.under what coditions?
sorry,under what conditions?
this is a question of theory of equations.
Do you have other assumptions on the cubic equation, like, how many solutions does it have, or other stuff like that?
no I don't know about the solutions but I know that this is a thoughtful question.
this is only verify and prove question.
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
but question is is true because my sir has solved it and i have lost that paper on which sir solved that.
I don't understand the question. You want us to solve for x? Is that what your asking?
one more thing about this quetion
all the roots of this cubic equations are real
Oh ok now the question makes more sense. lol/
Omg I'm sorry. My page got killed when I was typing all of that. :(
@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....
@KingGeorge any thoughts?
Sorry for asking this, but just to clarify, you're asking for conditions under which a will be less than 0 correct?
What do you think about taking the contrapositive?
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.
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|
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.
you both are good helper.