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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.

  • one year ago
  • one year ago

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  1. blockcolder
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    So your question is...?

    • one year ago
  2. maheshmeghwal9
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    my ques is why a<0.under what coditions?

    • one year ago
  3. maheshmeghwal9
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    sorry,under what conditions?

    • one year ago
  4. maheshmeghwal9
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    this is a question of theory of equations.

    • one year ago
  5. blockcolder
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    Do you have other assumptions on the cubic equation, like, how many solutions does it have, or other stuff like that?

    • one year ago
  6. maheshmeghwal9
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    no I don't know about the solutions but I know that this is a thoughtful question.

    • one year ago
  7. maheshmeghwal9
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    this is only verify and prove question.

    • one year ago
  8. veramath
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    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
  9. maheshmeghwal9
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    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
  10. freckles
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    I don't understand the question. You want us to solve for x? Is that what your asking?

    • one year ago
  11. cinar
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    http://www.sosmath.com/algebra/factor/fac11/fac11.html

    • one year ago
  12. maheshmeghwal9
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    what is this?

    • one year ago
  13. maheshmeghwal9
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    one more thing about this quetion

    • one year ago
  14. maheshmeghwal9
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    all the roots of this cubic equations are real

    • one year ago
  15. freckles
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    Oh ok now the question makes more sense. lol/

    • one year ago
  16. freckles
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    Omg I'm sorry. My page got killed when I was typing all of that. :(

    • one year ago
  17. maheshmeghwal9
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    np!

    • one year ago
  18. freckles
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    @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
  19. maheshmeghwal9
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    ok np. :(

    • one year ago
  20. freckles
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    i'm still thinking :(

    • one year ago
  21. freckles
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    @KingGeorge any thoughts?

    • one year ago
  22. KingGeorge
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    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
  23. freckles
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    What do you think about taking the contrapositive?

    • one year ago
  24. KingGeorge
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    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
  25. KingGeorge
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    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|

    • one year ago
  26. KingGeorge
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    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
  27. maheshmeghwal9
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    thanx!

    • one year ago
  28. KingGeorge
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    You're welcome.

    • one year ago
  29. freckles
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    yes thanks george :)

    • one year ago
  30. maheshmeghwal9
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    you both are good helper.

    • one year ago
  31. freckles
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    george is bester!

    • one year ago
  32. KingGeorge
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    Thank you guys :)

    • one year ago
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