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Math Help - I need a little help and a little confirmation please!

  1. #1
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    I need a little help and a little confirmation please!

    Hi again!
    I have a couple questions that I need help with.

    The first one I have the answer to but need a formula/proof/method/working/whatever you call it, to go with it.

    The question:
    The sum of a number and its reciprocal is 10/3. What is the number?

    My answer:
    3
    --------------------------------

    This next question confuses the (blank) out of me:

    The question
    ((x^2 + 5x + 4)/(x + 3)^2)/((x + 1)/(x^2 - 9))

    My answer:
    ???

    Thank you in advance for your help!
    Drakmord
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  2. #2
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    Quote Originally Posted by drakmord View Post
    Hi again!
    I have a couple questions that I need help with.

    [1] The first one I have the answer to but need a formula/proof/method/working/whatever you call it, to go with it.

    The question:
    The sum of a number and its reciprocal is 10/3. What is the number?

    My answer:
    3
    --------------------------------

    [2] This next question confuses the (blank) out of me:

    The question
    ((x^2 + 5x + 4)/(x + 3)^2)/((x + 1)/(x^2 - 9))

    My answer:
    ???

    Thank you in advance for your help!
    Drakmord
    to [1]:

    Let x denote the unknown number. Then you know:

    x + \dfrac1x = \dfrac{10}3

    Multiply through by x and rearrange the equation. You'll get:

    x^2-\dfrac{10}3 x + 1 = 0

    Solve this quadratic equation for x.

    to [2]

    Factorize each polynomal:

    \dfrac{\dfrac{x^2 + 5x + 4}{(x+3)^2}}{\dfrac{x+1}{x^2-9}}=\dfrac{\dfrac{(x+1)(x+4)}{(x+3)(x+3)}}{\dfrac{  x+1}{(x+3)(x-3)}} = \dfrac{(x+1)(x+4)}{(x+3)(x+3)} \cdot \dfrac{(x+3)(x-3)}{x+1}

    and now cancel.
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  3. #3
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    No "proof" needed, just write the equation out:

    The sum of a number and it's reciprocal:

    x+\frac{1}{x};

    Is equal to \frac{10}{3}:

    x+\frac{1}{x}=\frac{10}{3}.

    You can solve the above yes?

    As to your second problem, simply rewrite the problem by "flipping" the denominator, so instead of dividing by a rational, you are multiplying by its reciprocal.
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  4. #4
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    Ahhhh....
    I think I get it!
    So.....
    Would the answer to the question

    ((x^2 + 5x + 4)/(x + 3)^2)/((x + 1)/(x^2 - 9))

    be: x^2+x-12?
    ---------
    I think I can handle working though my first question now, thanks to both your help, ANDS! and earboth!

    Thank you for your help guys!

    Drakmord
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  5. #5
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    Quote Originally Posted by drakmord View Post
    Ahhhh....
    I think I get it!
    So.....
    Would the answer to the question

    ((x^2 + 5x + 4)/(x + 3)^2)/((x + 1)/(x^2 - 9))

    be: x^2+x-12? <<<<<<< unfortunately no
    ---------
    I think I can handle working though my first question now, thanks to both your help, ANDS! and earboth!

    Thank you for your help guys!

    Drakmord
    You forgot about the denominator (the numerator is OK!):

    <br />
\dfrac{(x+1)(x+4)}{(x+3)(x+3)} \cdot \dfrac{(x+3)(x-3)}{x+1} = \boxed{\dfrac{x^2+x-12}{x+3}}<br />
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  6. #6
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    Quote Originally Posted by earboth View Post
    You forgot about the denominator (the numerator is OK!):

    <br />
\dfrac{(x+1)(x+4)}{(x+3)(x+3)} \cdot \dfrac{(x+3)(x-3)}{x+1} = \boxed{\dfrac{x^2+x-12}{x+3}}<br />

    Ah.....oops! Thank you!
    How did I forget that!
    Thank you again for your help earboth!

    drakmord
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