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Thread: Factoring...1

  1. #1
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    Factoring...1

    What is the first step in factoring the following problem?

    (ax + b)^(-1/2) - root {(ax + b)/b}
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  2. #2
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    Re: Factoring...1

    Seems you'd start by using this rule: a^(-p) = 1 / a^p

    1 / (ax + b)^(1/2) - {(ax + b)/b}^(1/2)

    Remember that root n = n^(1/2)

    Also, if you continued the factoring process, I suggest
    setting ax + b = k, to get something simpler to work with:
    1 / k^(1/2) - (k/b)^(1/2)
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    Re: Factoring...1

    Quote Originally Posted by DenisB View Post
    Seems you'd start by using this rule: a^(-p) = 1 / a^p

    1 / (ax + b)^(1/2) - {(ax + b)/b}^(1/2)

    Remember that root n = n^(1/2)

    Also, if you continued the factoring process, I suggest
    setting ax + b = k, to get something simpler to work with:
    1 / k^(1/2) - (k/b)^(1/2)
    I'll work on this problem using k as a substitute for
    (ax + b) as you suggested and post my work this weekend.
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    Re: Factoring...1

    I tried but couldn't factor completely. Can someone work this out for me?
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    Re: Factoring...1

    So we start with (after substitution):

    1 / k^(1/2) - (k/b)^(1/2)

    right side: (k/b)^(1/2) = k^(1/2) / b^(1/2)
    so we now have:
    1 / k^(1/2) - k^(1/2) / b^(1/2)

    which is of style: 1/u - u/v

    I'll let you continue...
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    Re: Factoring...1

    Quote Originally Posted by DenisB View Post
    So we start with (after substitution):

    1 / k^(1/2) - (k/b)^(1/2)

    right side: (k/b)^(1/2) = k^(1/2) / b^(1/2)
    so we now have:
    1 / k^(1/2) - k^(1/2) / b^(1/2)

    which is of style: 1/u - u/v

    I'll let you continue...
    In place of v, you meant b.

    1/u - u/v = 1/u - u/b, right?

    This looks like a subtraction of two fractions.
    Do I proceed in terms of subtracting two fractions and then simplify some more?
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    Re: Factoring...1

    Quote Originally Posted by mathdad1965 View Post
    In place of v, you meant b.

    1/u - u/v = 1/u - u/b, right?
    No. I replaced k^(1/2) with u and b^(1/2 with v, to get 1/u - u/v

    Simplify that by multiplying by uv to get one fraction only.

    Then you're finished. You can substitute back in if you wish.
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  8. #8
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    Re: Factoring...1

    Quote Originally Posted by DenisB View Post
    No. I replaced k^(1/2) with u and b^(1/2 with v, to get 1/u - u/v

    Simplify that by multiplying by uv to get one fraction only.

    Then you're finished. You can substitute back in if you wish.
    Great. I can take it from here. I will post my work tomorrow.
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  9. #9
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    Re: Factoring...1

    1/u - 1/v becomes (v - u)/uv.

    What value for u and v must I substitute back in?
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    Re: Factoring...1

    Quote Originally Posted by mathdad1965 View Post
    1/u - 1/v becomes (v - u)/uv.

    What value for u and v must I substitute back in? Denis answered this in the quote box below, which was a couple of posts prior to this.
    That last denominator needs grouping symbols around it, such as:

    (v - u)/(uv)

    Quote Originally Posted by DenisB View Post
    No. I replaced k^(1/2) with u and b^(1/2)with v, to get 1/u - u/v
    $\dfrac{b^{1/2} - k^{1/2}}{k^{1/2}b^{1/2}} \ = $
    Last edited by greg1313; Jan 12th 2017 at 09:36 PM.
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    Re: Factoring...1

    Quote Originally Posted by greg1313 View Post
    That last denominator needs grouping symbols around it, such as:

    (v - u)/(uv)



    $\dfrac{b^{1/2} - k^{1/2}}{k^{1/2}b^{1/2}} \ = $
    So, do I now factor or is this the final answer?
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    Re: Factoring...1

    Quote Originally Posted by mathdad1965 View Post
    So, do I now factor or is this the final answer?
    Although k = ax + b in this method, I don't see a direct/handy route for showing a factor being taken out.

    Starting over:

    (ax + b)^(-1/2) - root {(ax + b)/b}

    $(ax + b)^{-1/2} - \dfrac{(ax + b)^{1/2}}{b^{1/2}} =$

    $(ax + b)^{-1/2} - (ax + b)^{1/2}b^{-1/2} =$

    $(ax + b)^{-1/2}{b^{-1/2}}[b^{1/2} - (ax + b)] = $

    $[b(ax + b)]^{-1/2}(b^{1/2} - ax - b)$


    After this post, I don't expect to post further on this thread.
    Last edited by greg1313; Jan 13th 2017 at 11:56 AM.
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  13. #13
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    Re: Factoring...1

    Quote Originally Posted by greg1313 View Post
    Although k = ax + b in this method, I don't see a direct/handy route for showing a factor being taken out.

    Starting over:

    (ax + b)^(-1/2) - root {(ax + b)/b}

    $(ax + b)^{-1/2} - \dfrac{(ax + b)^{1/2}}{b^{1/2}} =$

    $(ax + b)^{-1/2} - (ax + b)^{1/2}b^{-1/2} =$

    $(ax + b)^{-1/2}{b^{-1/2}}[b^{1/2} - (ax + b)] = $

    $[b(ax + b)]^{-1/2}(b^{1/2} - ax - b)$


    After this post, I don't expect to post further on this thread.
    I appreciate the breakdown and quick replies.
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