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Thread: Integral values

  1. #1
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    Integral values

    Hey guys,

    the task: For what integral values of k can 4x^2 + kx - 3 be factored?

    My answer:

    (4) (-3) = -12 Factors of 12= 1,12 2,6 and 3,4

    So are the possible values for k = +11, +4 and +1 or all negative?

    They should be positive right, because I am looking for k and the sign is positive in front of k.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Integral values

    You may either have:

    $\displaystyle (2x+1)(2x-3)$ or $\displaystyle (2x-1)(2x+3)$

    or

    $\displaystyle (4x+1)(x-3)$ or $\displaystyle (4x-1)(x+3)$
    Last edited by MarkFL; Oct 19th 2012 at 08:19 PM.
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  3. #3
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    Re: Integral values

    I think it should be

    (2x-1) (2x+3) or (2x+3)(2x-1)

    or

    (4x-1)(x+3) or (4x-3)(x+1)

    because

    (2x+1)(2x-3) = 4x^2 -4x-3 (in that case the value k would be negative) It should be 4x^2 + kx -3 (positive).
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Integral values

    I missed a pair:

    $\displaystyle (4x+3)(x-1)$ or $\displaystyle (4x-3)(x+1)$

    $\displaystyle k$ is just an arbitrary integer, it may either be positive or negative.
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  5. #5
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    Re: Integral values

    Okay, I got that.

    So, for what integral values of k can 4x^2 + kx + 3 be factored? (note +3, not -3 like in task 1).
    I would like to know your answer, because I got an example for 4x^2 + kx + 3 in my book. There is nothing about that k is just an arbitrary integer and it may be either positive or negative.
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Integral values

    When it is said "for what integral values of $\displaystyle k$" this implies that $\displaystyle k$ can be negative or positive, since the integers are {...-2,-1,0,1,2,...}.

    The possibilities here are:

    $\displaystyle (4x+1)(x+3)$ so $\displaystyle k=13$

    $\displaystyle (4x-1)(x-3)$ so $\displaystyle k=-13$

    $\displaystyle (4x+3)(x+1)$ so $\displaystyle k=7$

    $\displaystyle (4x-3)(x-1)$ so $\displaystyle k=-7$

    $\displaystyle (2x+1)(2x+3)$ so $\displaystyle k=8$

    $\displaystyle (2x-1)(2x-3)$ so $\displaystyle k=-8$
    Thanks from ford2008
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  7. #7
    Newbie Farenxdesign's Avatar
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    Re: Integral values

    I hate thiss..

    Fahrenheit Marketing

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  8. #8
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    Re: Integral values

    Hello, ford2008!

    For what integral values of $\displaystyle k$ can $\displaystyle 4x^2 + kx - 3$ be factored?

    My answer:

    (4) (-3) = -12 . . Factors of 12: (1,12), (2,6) and (3,4).

    So are the possible values are: .$\displaystyle k \;=\;\pm1,\,\pm4,\,\pm11$
    This is correct!

    My approach . . .

    We have: .$\displaystyle 4x^2 + kx - 3 \:=\:0$

    Quadratic Formula: .$\displaystyle x \;=\;\frac{-k \pm\sqrt{k^2 + 48}}{8}$

    To be factorable, the discriminant must be a square:
    . . $\displaystyle \text{that is: }\:k^2+48 \:=\:a^2\,\text{ for some integer }a.$

    This occurs when: .$\displaystyle k \;=\;\pm1,\,\pm4,\,\pm11$
    Thanks from ford2008
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