## maheshmeghwal9 Group Title 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. 2 years ago 2 years ago

1. blockcolder Group Title

2. maheshmeghwal9 Group Title

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

3. maheshmeghwal9 Group Title

sorry,under what conditions?

4. maheshmeghwal9 Group Title

this is a question of theory of equations.

5. blockcolder Group Title

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

6. maheshmeghwal9 Group Title

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

7. maheshmeghwal9 Group Title

this is only verify and prove question.

8. veramath Group Title

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

9. maheshmeghwal9 Group Title

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

10. freckles Group Title

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

11. cinar Group Title
12. maheshmeghwal9 Group Title

what is this?

13. maheshmeghwal9 Group Title

14. maheshmeghwal9 Group Title

all the roots of this cubic equations are real

15. freckles Group Title

Oh ok now the question makes more sense. lol/

16. freckles Group Title

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

17. maheshmeghwal9 Group Title

np!

18. freckles Group Title

@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....

19. maheshmeghwal9 Group Title

ok np. :(

20. freckles Group Title

i'm still thinking :(

21. freckles Group Title

@KingGeorge any thoughts?

22. KingGeorge Group Title

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

23. freckles Group Title

What do you think about taking the contrapositive?

24. KingGeorge Group Title

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.

25. KingGeorge Group Title

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|

26. KingGeorge Group Title

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.

27. maheshmeghwal9 Group Title

thanx!

28. KingGeorge Group Title

You're welcome.

29. freckles Group Title

yes thanks george :)

30. maheshmeghwal9 Group Title

you both are good helper.

31. freckles Group Title

george is bester!

32. KingGeorge Group Title

Thank you guys :)