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Math Help - Quadratic Equation Question

  1. #1
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    Question Quadratic Equation Question

    Quadratic Equation Question-quadratic-equation-question-2011-02-09.png

    The question is in the image.
    Let me know if you can't see it.
    I may add some questions regarding quadratic equations later..
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  2. #2
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    What does " x=\alpha,\beta" mean? Does it mean " x=\alpha or x=\beta", or " x=\alpha and x=\beta", or something else? Also, should it say ax^2+bx+cx or ax^2+bx+c?
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  3. #3
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    It means " x=\alpha and x=\beta".
    Sorry for the confusion.
    And it's supposed to say ax^2+bx+c, sorry.
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  4. #4
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    This information does not guarantee that ax^2+bx+c=a(x-\alpha)(x-\beta). For example, if a=b=c=1 and x=\alpha=\beta=0, then ax^2+bx+c=1, but a(x-\alpha)(x-\beta)=0.
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  5. #5
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    Again, I have to add & apologize.
    If ax^2+bx+c=0\ x=\alpha and x=\beta
    Then would factoring ax^2+bx+c=0 be a(x-\alpha)(x-\beta)?
    Last edited by Yuuki; February 9th 2011 at 01:35 AM.
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  6. #6
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    Sorry, I double posted.
    Any way I can delete this??
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  7. #7
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    Quote Originally Posted by Yuuki View Post
    Again, I have to add & apologize.
    If ax^2+bx+c=0\ x=\alpha and x=\beta
    Then would factoring ax^2+bx+c=0 be a(x-\alpha)(x-\beta)?
    Question should read x=\alpha or x=\beta, as x cannot be \alpha and \beta at the same time.

    In this case, \alpha and \beta are the roots of the quadratic equation, hence equation can be written:

    a(x-\alpha)(x-\beta)=0
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  8. #8
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    Please allow me to continue being a skeptic. I am having some mental block.

    Why is it that if \alpha, \beta are roots of ax^2+bx+c, then ax^2+bx+c=a(x-\alpha)(x-\beta)? The converse is obviously true. I mean that we are completely used to this, but since the question is so basic, the proof should be similarly low-level, based on simple arithmetic facts.
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  9. #9
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    Because the set of all polynomials, p(x), of degree two or less, such that p(\alpha)= 0 and p(\beta)= 0, forms a one-dimensional subspace of the vector space of all polynomials of degree two or less. And that is, in my opinion, based on very "low level arthmetic facts"- it's just tedious to say without using the "vector space" terminology.
    Last edited by HallsofIvy; February 9th 2011 at 02:57 AM.
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  10. #10
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    Let's consider only polynomials of degree 2 or less.

    Quote Originally Posted by HallsofIvy View Post
    Because the set of all polynomials, p(x), of degree two or less, such that p(\alpha)= 0 and p(\beta)= 0, form a one-dimensional subspace of the vector space of all polynomials of degree two or less. And that is, in my opinion, based on very "low level arthmetic facts"- it's just tedious to say without using the "vector space" terminology.
    This says that there exists a polynomials p_0(x) with the following property. For every p, p(\alpha)=\p(\beta)=0 iff there exists an a such that p(x)=ap_0(x).

    This fact it pretty similar to the original claim that p(\alpha)=p(\beta)=0 iff p(x)=a(x-\alpha)(x-\beta). The question is why it holds.

    The temptation is precisely to resort to more complex facts, but it is not clear that this does not involve circular reasoning. I would like to see a proof from simpler facts.
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  11. #11
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    Here is one way, inspired by Little Bézout's theorem. Suppose a\ne0 and \alpha and \beta are the only roots of ax^2+bx+c. Then, one can check directly that

    ax^2+bc+c=a(x+(b/a+\alpha))(x-\alpha)

    Indeed, the right-hand side expands to ax^2+bx-(a\alpha^2+b\alpha), but since \alpha is a root, -(a\alpha^2+b\alpha)=c.

    The roots of the left-hand side are \alpha and \beta, and the roots of the right-hand side are \alpha and -(b/a+\alpha). Therefore, \beta=-(b/a+\alpha).
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  12. #12
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    Thank you, emakarov, HallsofIvy, and Ithaka/

    I have another question.

    x^2+mx-6m^2=0
    In the above equation, solve m when x=-3
    I was able to solve the problem;
    When m=1, x=2 or x=-3
    When m=-\frac{3}{2}, x=\frac{9}{2} or x=-3

    I don't understand why x=-3 in both circumstances ( m=1 and =-\frac{3}{2})
    I know you can say that because the problem says "when x=-3", but I'm not satisfied with that answer.
    I think I may be content if I do the same problem with a different way of approaching it...
    Last edited by Yuuki; February 11th 2011 at 02:46 AM.
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  13. #13
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    Quote Originally Posted by Ithaka View Post
    Question should read x=\alpha or x=\beta, as x cannot be \alpha and \beta at the same time.

    In this case, \alpha and \beta are the roots of the quadratic equation, hence equation can be written:

    a(x-\alpha)(x-\beta)=0
    Completely off topic, but Ithica, you should use \displaystyle \pi in your siggie instead of \displaystyle \Pi, because \displaystyle \pi is the ratio of a circle's circumference to its diameter, while \displaystyle \Pi is used as the symbol for "product".
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  14. #14
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    Quote Originally Posted by Yuuki View Post
    m^2+m-6m^2=0
    In the above equation, solve m when x = -3
    This does not make much sense since there is no x in the equation. The solutions are m = 0 and m = 1/5.
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  15. #15
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    Quote Originally Posted by Prove It View Post
    Completely off topic, but Ithica, you should use \displaystyle \pi in your siggie instead of \displaystyle \Pi, because \displaystyle \pi is the ratio of a circle's circumference to its diameter, while \displaystyle \Pi is used as the symbol for "product".
    Yes, thank you very much. I know the difference between the 2 , it was a typing error made at the time I joined the forum and I overlooked it afterwards.
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