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Math Help - quadratic algebra

  1. #1
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    quadratic algebra

    hi all,

    the question posed is

    Given that x^2-14x+a = (x+b)^2 for all values of x, find the value of a and the value of b.

    is this correct?

    since

    (x+A)^2=x^2+2Ax+A^2

    x^2+2Ax=(x+A)^2-A^2

    so

    x^2-14x = (x+b)^2 - b^2 = x^2+2bx

    and solving for b

    b=\frac{-14x}{2x}=-7

    so

    x^2-14x+a=(x-7)^2

    and expanding the RHS

    a=49
    Last edited by sammy28; August 28th 2009 at 01:20 AM. Reason: incorrect sign
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  2. #2
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    Quote Originally Posted by sammy28 View Post
    hi all,

    the question posed is

    Given that x^2-14x+a = (x+b)^2 for all values of x, find the value of a and the value of b.

    is this correct?
    ...
    solving

    b=-7

    a=49
    That's the same as what I got.
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  3. #3
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    Your solution is correct, however there is a much easier way of getting to it:

    Simply look at the equation when x=0,-b (you can do this since you are told that it holds for any x

    Then, you get: x=0:

    0^2 -14*0 + a = (0 + b)^2 \Rightarrow a = b^2

    x=-b:
    (-b)^2 +14b + a = (b-b)^2 = 0 \Rightarrow b^2 + 14b + b^2 = 0 \Rightarrow
    \Rightarrow 2b^2 + 14b = 0 \Rightarrow 2b(b+7) = 0 \Rightarrow b = 0,-7

    This gives us two possible solutions, however a=b=0 is obviously wrong! so we are left with \boxed{b = -7, a = 49}
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  4. #4
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    Since two quadratic expressions,

    x^2+2bx+b^2

    and
    x^2-14x+a are identical,
    their coefficients will be in proportion.

    i.e. 1=-14/2b = (b^2)/a

    so 2b=-14 and a=b^2

    So  b=-7 , a=49
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  5. #5
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    thanks for the replies. its good to see it approached from different angles.

    defunkt i like your logic!

    thanks a lot
    sammy
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