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Mathematics 7 Online
OpenStudy (anonymous):

solve x^2+5x+6=0 by completing the square

OpenStudy (amistre64):

we need to add zero to this in order for it to be useful

OpenStudy (amistre64):

\[(x+a)(x+b)=x^2+(a+b)x+ab\] if b=a we have a perfect square (a complete square) \[(x+a)(x+a)=x^2+(a+a)x+aa\]\[(x+a)(x+a)=x^2+(2a)x+a^2\] notice that the last term of a completed square is half the middle term, squared \[(x+3)(x+3)=x^2+(a+a)x+aa\]\[(x+a)(x+a)=x^2+2(3)x+3^2\] and that is the key to completing a square

OpenStudy (amistre64):

.... copy paste failed :) last part should have been \[(x+3)(x+3)=x^2+2(3)x+3^2\]

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