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Thread: 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 "$\displaystyle x=\alpha,\beta$" mean? Does it mean "$\displaystyle x=\alpha$ or $\displaystyle x=\beta$", or "$\displaystyle x=\alpha$ and $\displaystyle x=\beta$", 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 "$\displaystyle x=\alpha$ and $\displaystyle x=\beta$".
    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 $\displaystyle ax^2+bx+c=a(x-\alpha)(x-\beta)$. For example, if $\displaystyle a=b=c=1$ and $\displaystyle x=\alpha=\beta=0$, then $\displaystyle ax^2+bx+c=1$, but $\displaystyle a(x-\alpha)(x-\beta)=0$.
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  5. #5
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    Again, I have to add & apologize.
    If $\displaystyle ax^2+bx+c=0\ x=\alpha$ and $\displaystyle x=\beta$
    Then would factoring $\displaystyle ax^2+bx+c=0$ be $\displaystyle a(x-\alpha)(x-\beta)$?
    Last edited by Yuuki; Feb 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 $\displaystyle ax^2+bx+c=0\ x=\alpha$ and $\displaystyle x=\beta$
    Then would factoring $\displaystyle ax^2+bx+c=0$ be $\displaystyle a(x-\alpha)(x-\beta)$?
    Question should read $\displaystyle x=\alpha$ or $\displaystyle x=\beta$, as x cannot be $\displaystyle \alpha$ and $\displaystyle \beta$ at the same time.

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

    $\displaystyle a(x-\alpha)(x-\beta)=0$
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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 $\displaystyle \alpha$, $\displaystyle \beta$ are roots of $\displaystyle ax^2+bx+c$, then $\displaystyle ax^2+bx+c=a(x-\alpha)(x-\beta)$? 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 $\displaystyle p(\alpha)= 0$ and $\displaystyle p(\beta)= 0$, 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; Feb 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 $\displaystyle p(\alpha)= 0$ and $\displaystyle p(\beta)= 0$, 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 $\displaystyle p_0(x)$ with the following property. For every $\displaystyle p$, $\displaystyle p(\alpha)=\p(\beta)=0$ iff there exists an $\displaystyle a$ such that $\displaystyle p(x)=ap_0(x)$.

    This fact it pretty similar to the original claim that $\displaystyle p(\alpha)=p(\beta)=0$ iff $\displaystyle p(x)=a(x-\alpha)(x-\beta)$. 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 $\displaystyle a\ne0$ and $\displaystyle \alpha$ and $\displaystyle \beta$ are the only roots of $\displaystyle ax^2+bx+c$. Then, one can check directly that

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

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

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

    I have another question.

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

    I don't understand why $\displaystyle x=-3$ in both circumstances ($\displaystyle m=1$ and $\displaystyle =-\frac{3}{2}$)
    I know you can say that because the problem says "when $\displaystyle x=-3$", 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; Feb 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 $\displaystyle x=\alpha$ or $\displaystyle x=\beta$, as x cannot be $\displaystyle \alpha$ and $\displaystyle \beta$ at the same time.

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

    $\displaystyle a(x-\alpha)(x-\beta)=0$
    Completely off topic, but Ithica, you should use $\displaystyle \displaystyle \pi$ in your siggie instead of $\displaystyle \displaystyle \Pi$, because $\displaystyle \displaystyle \pi$ is the ratio of a circle's circumference to its diameter, while $\displaystyle \displaystyle \Pi$ is used as the symbol for "product".
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  14. #14
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    Quote Originally Posted by Yuuki View Post
    $\displaystyle m^2+m-6m^2=0$
    In the above equation, solve $\displaystyle m$ when $\displaystyle x = -3$
    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 $\displaystyle \displaystyle \pi$ in your siggie instead of $\displaystyle \displaystyle \Pi$, because $\displaystyle \displaystyle \pi$ is the ratio of a circle's circumference to its diameter, while $\displaystyle \displaystyle \Pi$ 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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