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maheshmeghwal9
 3 years 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.
maheshmeghwal9
 3 years 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.

This Question is Closed

blockcolder
 3 years ago
Best ResponseYou've already chosen the best response.0So your question is...?

maheshmeghwal9
 3 years ago
Best ResponseYou've already chosen the best response.0my ques is why a<0.under what coditions?

maheshmeghwal9
 3 years ago
Best ResponseYou've already chosen the best response.0sorry,under what conditions?

maheshmeghwal9
 3 years ago
Best ResponseYou've already chosen the best response.0this is a question of theory of equations.

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

maheshmeghwal9
 3 years ago
Best ResponseYou've already chosen the best response.0no I don't know about the solutions but I know that this is a thoughtful question.

maheshmeghwal9
 3 years ago
Best ResponseYou've already chosen the best response.0this is only verify and prove question.

veramath
 3 years ago
Best ResponseYou've already chosen the best response.0The 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
 3 years ago
Best ResponseYou've already chosen the best response.0but question is is true because my sir has solved it and i have lost that paper on which sir solved that.

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

maheshmeghwal9
 3 years ago
Best ResponseYou've already chosen the best response.0one more thing about this quetion

maheshmeghwal9
 3 years ago
Best ResponseYou've already chosen the best response.0all the roots of this cubic equations are real

freckles
 3 years ago
Best ResponseYou've already chosen the best response.1Oh ok now the question makes more sense. lol/

freckles
 3 years ago
Best ResponseYou've already chosen the best response.1Omg I'm sorry. My page got killed when I was typing all of that. :(

freckles
 3 years ago
Best 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....

freckles
 3 years ago
Best ResponseYou've already chosen the best response.1@KingGeorge any thoughts?

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

freckles
 3 years ago
Best ResponseYou've already chosen the best response.1What do you think about taking the contrapositive?

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

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.2This 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
 3 years ago
Best ResponseYou've already chosen the best response.0you both are good helper.
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