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Equation help

  1. Oct 31, 2004 #1
    If (x + a)² + b = x² - 6x + 13
    find the values of a and b

    there are 3 unknowns and this confuses me... please help
     
  2. jcsd
  3. Oct 31, 2004 #2

    shmoe

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    You have a polynomial on the left hand side and a polynomial on the right hand side. Two polynomials are equal if and only if their coefficients match.

    eg.if [tex]2x^2-3x+4=ax^2+bx+c[/tex] then we must have [tex]a=2,b=-3,c=4[/tex]

    For your question, try expanding the [tex](x+a)^2[/tex] part and equating coefficients.
     
  4. Oct 31, 2004 #3
    x² + 2ax + a² + b = x² - 6x + 13
    the x² cancel out
    a² + 2ax + b = 13 - 6x

    then I get stuck
     
  5. Oct 31, 2004 #4

    shmoe

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    Collect like terms (according to power of "x"). Your equation becomes:

    [tex](2a)x+(a^2+b)=-6x+13[/tex]

    Switching the order of the left hand side was just to make things match up nicer. Remember what I said about the coefficients of equal polynomials.

    [tex]2a=??[/tex]
    [tex]a^2+b=??[/tex]

    Can you fill in the ??
     
  6. Oct 31, 2004 #5

    arildno

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    Write your last equation as:
    (2a+6)x=13-a^2-b
    Since this equation shall hold for ALL choices of x, in particular it must hold for x=0
    which means you get the equations:
    0=13-a^2-b
    And:
    2a+6=0
     
  7. Oct 31, 2004 #6
    so:
    a=-3
    and:
    0=16^2-b
    b=2
     
  8. Oct 31, 2004 #7

    arildno

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    How do you get that flawed b-value?????
     
  9. Oct 31, 2004 #8
    I see, I did:
    16^(2-b)
    I should do:
    13-(3^2)-b
    b=4
     
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