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Math Help - Completing the square #2

  1. #1
    Senior Member Mukilab's Avatar
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    Completing the square #2

    9x^2+12x+7

    Can I do

    (9x+12x+36)-29?
    Last edited by mr fantastic; May 29th 2010 at 06:08 AM. Reason: Edited post title.
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  2. #2
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    Quote Originally Posted by Mukilab View Post
    9x^2+12x+7

    Can I do

    (9x\color{red}^2\color{black}+12x+36)-29?
    Not quite,

    9x^2+12x+7=(3x)^2+12x+7=(3x)^2+2(6x)+7= (3x)^2+2[3(2x)]+7=(3x)^2+2[3(2x)]+2^2-2^2+7=(3x+2)^2+7-4

    Alternatively,

    9\left(x^2+\frac{12}{9}x\right)+7=9\left(x^2+2\lef  t[\frac{6}{9}x\right]+\left[\frac{6}{9}\right]^2\right)+7-9\left(\frac{6}{9}\right)^2 etc
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  3. #3
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    Quote Originally Posted by Mukilab View Post
    9x^2+12x+7

    Can I do

    (9x+12x+36)-29?
    9x^2 + 2*3*2*x + 4 + 3

    (3x + 2)^2 + 3
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  4. #4
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    Quote Originally Posted by Mukilab View Post
    9x^2+12x+7

    Can I do

    (9x+12x+36)-29?
    No, the coefficient of x^2 needs to be 1.

    So what you need to do is...

    9x^2 + 12x + 7 = 9\left(x^2 + \frac{4}{3}x + \frac{7}{9}\right)

     = 9\left[x^2 + \frac{4}{3}x + \left(\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2 + \frac{7}{9}\right]

     = 9\left[\left(x + \frac{2}{3}\right)^2 - \frac{4}{9} + \frac{7}{9}\right]

     = 9\left[\left(x + \frac{2}{3}\right)^2 + \frac{3}{9}\right]

     = 9\left[\left(x + \frac{2}{3}\right)^2 + \frac{1}{3}\right]

     = 9\left(x + \frac{2}{3}\right)^2 + 3.
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