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Find the Values of the constants in the following indentities

  1. Aug 30, 2005 #1
    Find the Values of the constants in the following indentities.

    A(X^2 -1) + B(X - 1) + C = (3x -1)(x +1)

    AX^2 - A + Bx - B + C = 3x^2 + 3x -x -1

    Ok, Is this correct so far?

    AX^2 = 3x^2 :. A = 3
    Is correct ?

    Bx = 3x = b = 3 ?
    C = -1 ?

    I'm sure this is wrong, I just need help explaining why and correcting my errors. It seems i need to simplify it more before comparing the co-efficients.

    Thanks in advanced guys,
    Probably all find this easy :blushing:
     
  2. jcsd
  3. Aug 30, 2005 #2

    TD

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    So, you start with

    [tex]A\left( {x^2 - 1} \right) + B\left( {x - 1} \right) + C = \left( {3x - 1} \right)\left( {x + 1} \right)[/tex]

    You worked out both sides

    [tex]Ax^2 - A + Bx - B + C = 3x^2 + 3x - x - 1[/tex]

    Now, before you continue, rearrange both sides so you group per power of x

    [tex]Ax^2 + Bx - A - B + C = 3x^2 + 2x - 1[/tex]

    Now compare the coefficients of the equal powers of x of both sides. Remember that the all the constants are the coefficients of x^0.
     
    Last edited: Aug 30, 2005
  4. Aug 30, 2005 #3
    Just a small error. You forgot to simplify [itex]3x-x[/itex] to [itex]2x[/itex].

    [tex]A(x^2-1) + B(x-1) + C = (3x-1)(x+1)[/tex]
    <=>
    [tex]Ax^2-A+Bx-B+C = 3x^2+3x-x-1 = 3x^2+2x-1 [/tex]
    <=>
    [tex]Ax^2+Bx-A-B+C = 3x^2+2x-1 [/tex]

    Now try again.
     
  5. Aug 30, 2005 #4
    TD: It's [itex]+2x[/itex], not [itex]-2x[/itex] as in your post.
     
  6. Aug 30, 2005 #5

    TD

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    Right, I'll correct :)
     
  7. Aug 30, 2005 #6
    Bingo, gee thats easy ... but i always think that, then ill try a new one and get confused again.

    These books im learning from arn't to clear.

    Thanks guys,
    Ax^2 + Bx - A - B +C = 3x^2 + 2x - 1
    A = 3
    B = 2
    -A -B + C = -1
    -3 -2 + C = - 1
    C = 4
    -5 + 4 = - 1

    Excellent.
     
  8. Aug 30, 2005 #7

    TD

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    Seems correct :smile:
     
  9. Aug 30, 2005 #8
    One more question, a new identity but basically expanding with fractions.

    (X^2 - A/X^2) (X^2 - A/X^2)

    = X^4 - A/X^4 - A/X^4 + A^2/X^4

    X^4 - 2A/x^8 + A^2/X^4 ???

    But is the 2A/X^8 Correct? Sure its wrong????
     
  10. Aug 30, 2005 #9
    [itex](a-b)(a-b) = (a-b)^2 = a^2 - 2ab + b^2 [/itex]

    So [itex]-2x^2\frac{A}{x^2}=-2A[/tex] not [itex]2A/X^8[/itex]
     
  11. Aug 30, 2005 #10
    How do you make it look like that?

    My layouits are unclear so i think you've given the wrong explanation(My fault).

    Wondering how its 2A/Nothing

    Rather than
    2A/x^8

    From -A/x^4 -A/X^4

    SOmehow they cancel eachother? (The bottom bits) leaving 2A/ Nothing.
     
  12. Aug 30, 2005 #11
    I assume you know that [itex](a-b)(a-b) = (a-b)^2 = a^2 - 2ab + b^2[/itex] (*)

    So:
    [tex](x^2 - \frac{A}{x^2}) (x^2 - \frac{A}{x^2}) = (x^2 - \frac{A}{x^2})^2 = (*) = (x^2)^2 -2x^2\frac{A}{x^2} + (\frac{A}{x^2})^2 = x^4 - 2A + \frac{A^2}{x^4}[/tex]

    What I'm saying is that:
    [tex]-2x^2\frac{A}{x^2} = \frac{-2Ax^2}{x^2} = -2A[/tex]
    since the [itex]x^2[/itex] in the numerator and denominator cancels out.
     
  13. Aug 30, 2005 #12
    Wow thats confusing!

    So can you quickly type out[itex] (X^2 - A/x^2)^2 [/itex] Please

    Because i get:

    [itex]+x^4 - A/x^4 - A/x^4 + A^2/x^4[/itex]
    Just want to know how X^4 cancel eachother out? Or have i done this basic step wrong in the first place.
     
  14. Aug 30, 2005 #13
    Yes, you seem to believe that
    [tex]x^2\frac{A}{x^2} = \frac{A}{x^4}[/tex]
    which is not true.

    In fact [itex]x^2\frac{A}{x^2} = A[/itex]

    So
    [tex](x^2-A/x^2)(x^2-A/x^2) = x^4 - A - A + A^2/X^4[/tex]


    [Edit]: Missed one +-sign.
     
    Last edited: Aug 30, 2005
  15. Aug 30, 2005 #14
    So if X^2 is below something(Demoninator i think you call it?) and its multiplyed by itself, It cancels eachother out?

    Just trying to get my head around this, Quite confused lol.

    Basically [itex]X^2 [Multiplied by ] A/X^2[/itex] Cancels X^2 leaving A/ Nothing.
     
  16. Aug 30, 2005 #15
    That's right. Just take a look at this trivial example:

    [tex]1 = \frac{1}{2} + \frac{1}{2} = (definition) = 2*\frac{1}{2} = 1[/tex]
     
  17. Aug 30, 2005 #16

    TD

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    A little bit of terminology:

    - we call [itex]\frac{A}{B}[/itex] a fraction
    - A is the nominator
    - B is the denominator

    Perhaps that makes it easier to follow :smile:
     
  18. Aug 30, 2005 #17

    HallsofIvy

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    Well, not A over Nothing! That would be A/0 which makes no sense.
    The cancelling leaves 1. x2(A/x2= A/1= A.
     
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