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Thread: need help with doubt about exercise of polynomials (algebraic expressions)

  1. #1

    Question need help with doubt about exercise of polynomials (algebraic expressions)

    Hello to all the community , I'm new to this forum

    I apologize if Iwrite something that is not understood , I will be very concrete.

    The exercise is as follows.

    calculate and indicate that values of X is defined the ratio.



    My resolution

    First thing I did was multiply polynomials crossed (x-3 ) * (8x + 16x^2)

    and ( 4x ) * ( x
    ^2 - 9 )

    and then re-divide the remaining fraction of polynomials but I realized that Ido not get anything.

    Then I can not understand the part where it says

    "for values of X is defined the ratio"

    Please if anyone can help I would appreciate it !

    Best of luck to everyone!



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  2. #2
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    The bit about "indicate that values of X is defined the ratio" seems like it was poorly translated from another language. I don't understand it any better than you do, so I'd recommend asking your teacher for clarification, as only he/she can tell you what the worksheet is really asking. Until I know more about the problem, I don't feel comfortable conjecturing about what it might mean.

  3. #3
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by au6us7o View Post
    I can not understand the part where it says

    "for values of X is defined the ratio"
    The directions are saying that, assuming the two sides are not always equal, find x-values for which the two sides are equal. For instance, in the following ratio:

    . . . . .[tex]\dfrac{x}{3}\, :\, \dfrac{2}{1}[/tex]

    This is true exactly and only for:

    . . . . .[tex]1\, \cdot\, x\, =\, 2\, \cdot\, 3[/tex]

    . . . . .[tex]x\, =\, 6[/tex]

    However, in the following case:

    . . . . .[tex]\dfrac{x}{3}\, :\, \dfrac{9x}{27}[/tex]

    ...the ratio is true for all values of the variable:

    . . . . .[tex]27x\, =\, 3(9x)[/tex]

    . . . . .[tex]27x = 27x[/tex]

    This is trivially true if the variable equals zero, so assume not, and divide through:

    . . . . .[tex]27\, =\, 27[/tex]

    So the value of the variable in this particular case is irrelevant; the ratio is true "for all values of" the variable.

    In your case, you had:

    . . . . .[tex]\dfrac{x\, -\, 3}{4x}\, :\, \dfrac{x^2\, -\, 9}{8x\, +\, 16x^2}[/tex]

    . . . . .[tex]\dfrac{x\, -\, 3}{4x}\, :\, \dfrac{(x\, -\, 3)(x\, +\, 3)}{(4x)(2\, +\, 4x)}[/tex]

    So what must be proportional to 1? Doing the cross-multiplication and solving the resulting equation, what do you get?

    Quote Originally Posted by au6us7o View Post
    calculate and indicate that values of X is defined the ratio.

    [tex]\mbox{a) }\, \dfrac{x\, -\, 3}{4x}\, :\, \dfrac{x^2\, -\, 9}{8x\, +\, 16x^2}[/tex]

    First thing I did was multiply polynomials crossed:

    [tex](x\, -\, 3)\, (8x\, +\, 16x^2)\, =\, (4x)\, (x^2\, -\, 9)[/tex]

    and then re-divide the remaining fraction of polynomials but I realized that I do not get anything.
    What did you do when you "re-divided"? What is the "remaining fraction of polynomials"? What did you get, that you concluded was "not ... anything"?

    Please be complete. Thank you!

  4. #4
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    simplify?

    I believe the problem wants you to simplify (reduce) the proportion.
    Perhaps:

    1 : (x+3)/(4x+2) when (x-3)#0 and x#0

    or

    (4x+2) : (x+3) when x#-1/2, x#3 and x#0
    Last edited by Bob Brown MSEE; 11-24-2015 at 03:10 PM.
    "What happens in the event horizon, stays in the event horizon" -- Bob Brown (grandpa Bob)

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