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Completing the Square

  1. Dec 16, 2004 #1


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    I have only just recently started this topic ... (3 topic in AS maths for me)

    I have 1 question with two parts.I just can't seem to get the answer!

    "Express, in the form [tex](px+q)^2 + r [/tex] whereby p > 0"

    a) [tex] 16x^2 -8x +11[/tex]
    b) [tex] 9x^2 +3x +1[/tex]

    I don't seem to find a problem doing any of these when the coefficient of X squared is 1, but when it is bigger than 1, it causes me problems!

    I would be gratefull if some once could at least help me through the first 1, so i can understand the method.Thanks.

  2. jcsd
  3. Dec 16, 2004 #2


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    I'll solve a) and let u take b).
    a)[tex]16x^{2}=(4x)^{2} [/tex]
    [tex] 16x^{2}-8x=(4x)^{2}-2\cdot 4x [/tex]
    [tex] 16x^2 -8x +11=(4x-1)^{2}+10 [/tex]

  4. Dec 16, 2004 #3


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    A naive way of doing this is to write:

    [tex] 16x^2 -8x +11=(px+q)^2 + r [/tex]

    and expand the right hand side (multiple out the squared part and collect powers of x). Now match the coefficients to solve for p, q, and r.
  5. Dec 16, 2004 #4


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    Thanks both for your help!
  6. Dec 17, 2004 #5
    I have never seen your method before. I cannot say it makes sense to me. care to explain it?
  7. Dec 17, 2004 #6


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    It's not a bigdeal.
    Let's pick an arbitrary polynom of degree 2:[tex] ax^{2}+bx+c [/tex].
    U wanna put in the form [tex] (px+q)^{2}+r [/tex].
    The direct method is to equal the two expressions and identify the coefficients of the powers of "x".That what Shmoe said.
    I found another method which can be thought of being intuitive,and sometimes useful as well.
    Take the square:[tex] ax^{2}+bx [/tex].It can be put like:
    [tex] (\sqrt{a} x)^{2}+2\sqrt{a}\frac{b}{2\sqrt{a}} x+\frac{b^{2}}{4a}-\frac{b^{2}}{4a} [/tex] okay??
    You restrain the square and add "c" in both sides to get:
    [tex]ax^{2}+bx+c =(\sqrt{a} x +\frac{b}{2\sqrt{a}})^{2} +(c-\frac{b^{2}}{4a}) [/tex]

    And u can easily find (p,q,r).

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