Completing Squares

[matt826]

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.

 

[Denis]

Re: Road... Quadratic Formula

Please start your own thread. Not "nice" to tailgate on somebody else's :shock:

 

[Deleted member 4993]

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

 

[BigGlenntheHeavy]

\(\displaystyle 4x^2+16x+12 \ = \ 0\)

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

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

\(\displaystyle Check: \ 4[x^2+4x+4-1] \ = \ 4[x^2+4x+3] \ = \ 4x^2+16x+12\)

 

[Denis]

\(\displaystyle 4x^2+16x+12 \ = \ 0\)
\(\displaystyle 4[x^2+4x+3] \ = \ 0. \ pull \ out \ common \ factors \ first \ to \ avoid \ any \ later \ angst.\)

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 = ?