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Thread: [SOLVED] calc? ulate this fractions

  1. #1
    asdfghjkl
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    [SOLVED] calc? ulate this fractions

    For some of my math i just cant figure out what i have to do just to start
    Any help or review to show me how to do some would be great?
    here is a link to the questions
    http://www.math.unb.ca/ready/paper/
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by asdfghjkl
    For some of my math i just cant figure out what i have to do just to start
    Any help or review to show me how to do some would be great?
    here is a link to the questions
    http://www.math.unb.ca/ready/paper/
    There is a lot here, and it would be a mistake for us to do them all for you.
    But I see no reason for not showing you how to do a selection of these.

    1a. Simplify: $\displaystyle \frac{x^3-9x}{x^2-7x+12}$.

    First if $\displaystyle x$ is a common factor of the top or bottom of a
    rational expression like this, factor it out:

    $\displaystyle \frac{x^3-9x}{x^2-7x+12} = \frac{x(x^2-9)}{x^2-7x+12} $.

    Now look for terms which are the difference of two squares as we know
    that $\displaystyle a^2+b^2=(a+b)(a-b)$, here we have such a term, so:

    $\displaystyle
    \frac{x^3-9x}{x^2-7x+12} = \frac{x(x^2-9)}{x^2-7x+12} $
    $\displaystyle
    =\frac{x(x+3)(x-3)}{x^2-7x+12}
    $

    Now we check if we can cancel any terms, in this case we check if:

    $\displaystyle x^2-7x+12$,

    has one or more of $\displaystyle x$, $\displaystyle x+3$ or $\displaystyle x-3$
    as factors. The first obviously is not a factor, if the second was a factor
    then substituting $\displaystyle x=-3$ into the expression would simplify to 0,
    it does not so $\displaystyle x+3$ id not a factor of $\displaystyle x^2-7x+12$.
    Now if the last candidate is a factor substituting $\displaystyle x=3$ into the
    expressions would simplify to zero, which in this case it does, and by trial
    and error we find:

    $\displaystyle x^2-7x+12=(x-3)(x-4)$, so now we have:

    $\displaystyle
    \frac{x^3-9x}{x^2-7x+12} = \frac{x(x^2-9)}{x^2-7x+12} $
    $\displaystyle
    =\frac{x(x+3)(x-3)}{x^2-7x+12}
    $$\displaystyle
    =\frac{x(x+3)(x-3)}{(x-3)(x-4)}
    $
    .

    Finally cancel common terms to get:

    $\displaystyle
    \frac{x^3-9x}{x^2-7x+12} = \frac{x(x^2-9)}{x^2-7x+12} $
    $\displaystyle
    =\frac{x(x+3)(x-3)}{x^2-7x+12}
    $$\displaystyle
    =\frac{x(x+3)(x-3)}{(x-3)(x-4)}
    $$\displaystyle
    =\frac{x(x+3)}{x-4}
    $
    .

    RonL
    Last edited by CaptainBlack; Aug 28th 2006 at 08:41 AM.
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  3. #3
    Grand Panjandrum
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    Don't make duplicate posts in a number of forums, it confuses and wastes the
    time of the helpers who cannot tell if a question has been answered elsewhere.

    Other thread deleted.

    RonL
    Last edited by CaptainBlack; Aug 28th 2006 at 09:00 AM.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by asdfghjkl
    For some of my math i just cant figure out what i have to do just to start
    Any help or review to show me how to do some would be great?
    here is a link to the questions
    http://www.math.unb.ca/ready/paper/
    2a. Rationalise the denominator of:

    $\displaystyle
    \frac{2}{\sqrt{3}+\sqrt{2}}\ \ \ \ ...(1)
    $

    Now remember what we said earlier about the difference od two squares:

    $\displaystyle
    a^2-b^2=(a+b)(a-b)
    $

    If we put $\displaystyle a=\sqrt{3}$ and $\displaystyle b=\sqrt{2}$, we have:

    $\displaystyle
    3-2=(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})
    $

    So if we now multiply the top and bottom of $\displaystyle (1)$ by $\displaystyle \sqrt{3}-\sqrt{2}$ we don't change its value so:

    $\displaystyle
    \frac{2}{\sqrt{3}+\sqrt{2}}=\frac{2(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})} $$\displaystyle
    =\frac{2\sqrt{3}-2\sqrt{2}}{3-2}=2\sqrt{3}-2\sqrt{2}
    $

    RonL
    Last edited by CaptainBlack; Aug 29th 2006 at 08:05 AM.
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  5. #5
    Grand Panjandrum
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    4a. Solve for x (without using a claculator)

    $\displaystyle
    5^{x+1}=25
    $

    $\displaystyle 25=5\times 5=5^2$, so:

    $\displaystyle
    5^{x+1}=5^2
    $.

    Therefore equating exponents $\displaystyle x+1=2$, and so $\displaystyle x=1$.

    Can also be done by taking logs to the base 5:

    $\displaystyle
    (x+1)\log_5(5)=2\log_5(5)
    $

    but $\displaystyle \log_5(5)=1$, so we are back to where we were, and $\displaystyle x=1$

    RonL
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  6. #6
    Grand Panjandrum
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    5a. Simplify:

    $\displaystyle
    \log_2(5)+\log_2(x^2-1)-\log_2(x-1)
    $.

    Here the first things we need are some of the basic laws of logarithms:

    $\displaystyle \log_a(u)+\log_s(v)=\log_a(u\times v)$,

    and:

    $\displaystyle \log_a(u)-\log_s(v)=\log_a(u/v)$.

    Now:

    $\displaystyle
    \log_2(5)+\log_2(x^2-1)$$\displaystyle -\log_2(x-1)=(\log_2(5)+\log_2(x^2-1))-\log_2(x-1)
    $
    $\displaystyle
    =\log_2(5(x^2-1))-\log_2(x-1)
    $$\displaystyle
    =\log \left(\frac{5(x^2-1}{x-1}\right)
    $

    Finally we need to recall how to factorise the difference of two squares to get:

    $\displaystyle
    =\log \left(\frac{5(x-1)(x+1)}{x-1}\right)=\log(5(x+1))
    $

    RonL
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