Completing Squares

matt826

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May 6, 2010
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Help please. Use the technique of completing the square to transform the quadratic equation 4x[sup:15e5u0fi]2[/sup:15e5u0fi] + 16x + 12 = 0 into the form (x + c)[sup:15e5u0fi]2[/sup:15e5u0fi] = a.
 
Re: Road... Quadratic Formula

Please start your own thread. Not "nice" to tailgate on somebody else's :shock:
 
matt826 said:
Help please. Use the technique of completing the square to transform the quadratic equation 4x[sup:sd17d7l9]2[/sup:sd17d7l9] + 16x + 12 = 0 into the form (x + c)[sup:sd17d7l9]2[/sup:sd17d7l9] = a.

Please share with us your work, indicating exactly where you are stuck - so that we may know where to begin to help you
 
4x2+16x+12 = 0\displaystyle 4x^2+16x+12 \ = \ 0

4[x2+4x+3] = 0. pull out common factors first to avoid any later angst.\displaystyle 4[x^2+4x+3] \ = \ 0. \ pull \ out \ common \ factors \ first \ to \ avoid \ any \ later \ angst.

4[(x+2)21] = 0, Done.\displaystyle 4[(x+2)^2-1] \ = \ 0, \ Done.

Check: 4[x2+4x+41] = 4[x2+4x+3] = 4x2+16x+12\displaystyle Check: \ 4[x^2+4x+4-1] \ = \ 4[x^2+4x+3] \ = \ 4x^2+16x+12
 
BigGlenntheHeavy said:
4x2+16x+12 = 0\displaystyle 4x^2+16x+12 \ = \ 0
4[x2+4x+3] = 0. pull out common factors first to avoid any later angst.\displaystyle 4[x^2+4x+3] \ = \ 0. \ pull \ out \ common \ factors \ first \ to \ avoid \ any \ later \ angst.
Just divide by 4, easier: x^2 + 4x + 3 = 0
Since (x+2)^2 = x^2 + 4x + 4, then x^2 + 4x + 3 = (x+2)^2 - 1 = ?
 
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