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maheshmeghwal9

  • 2 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.

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  1. blockcolder
    • 2 years ago
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    So your question is...?

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

  3. maheshmeghwal9
    • 2 years ago
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    sorry,under what conditions?

  4. maheshmeghwal9
    • 2 years ago
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    this is a question of theory of equations.

  5. blockcolder
    • 2 years ago
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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?

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

  7. maheshmeghwal9
    • 2 years ago
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    this is only verify and prove question.

  8. veramath
    • 2 years ago
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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

  9. maheshmeghwal9
    • 2 years ago
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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.

  10. freckles
    • 2 years ago
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    I don't understand the question. You want us to solve for x? Is that what your asking?

  11. cinar
    • 2 years ago
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    http://www.sosmath.com/algebra/factor/fac11/fac11.html

  12. maheshmeghwal9
    • 2 years ago
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    what is this?

  13. maheshmeghwal9
    • 2 years ago
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    one more thing about this quetion

  14. maheshmeghwal9
    • 2 years ago
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    all the roots of this cubic equations are real

  15. freckles
    • 2 years ago
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    Oh ok now the question makes more sense. lol/

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

  17. maheshmeghwal9
    • 2 years ago
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    np!

  18. freckles
    • 2 years ago
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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....

  19. maheshmeghwal9
    • 2 years ago
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    ok np. :(

  20. freckles
    • 2 years ago
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    i'm still thinking :(

  21. freckles
    • 2 years ago
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    @KingGeorge any thoughts?

  22. KingGeorge
    • 2 years ago
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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?

  23. freckles
    • 2 years ago
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    What do you think about taking the contrapositive?

  24. KingGeorge
    • 2 years ago
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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.

  25. KingGeorge
    • 2 years ago
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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|

  26. KingGeorge
    • 2 years ago
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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.

  27. maheshmeghwal9
    • 2 years ago
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    thanx!

  28. KingGeorge
    • 2 years ago
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    You're welcome.

  29. freckles
    • 2 years ago
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    yes thanks george :)

  30. maheshmeghwal9
    • 2 years ago
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    you both are good helper.

  31. freckles
    • 2 years ago
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    george is bester!

  32. KingGeorge
    • 2 years ago
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    Thank you guys :)

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