math
posted by gagan .
how to complete the square for
2x^2+3x1?
can you plz help me, and any help is greatly appricieted.
thnx a lot for your effort.

step1
always factor out the coefficient of the x^2 term, unless it is already 1
this guarantees that it will be just x^2
2x^2+3x1
= 2[x^2  3/2 + ...]  1
now take 1/2 of the coefficient of the x term, square that result, and then add and subtract it
1/2 of 3/2 is 3/4, which when squared is 9/16
= 2[x^2  3/2 + 9/16  9/16]  1
your first 3 terms inside the bracket are your "perfect square"
= 2[x  3/4)^2  9/16]  1
now multiply through by the 2
= 2(x  3/4)^2 + 9/8  1
simplify the end part
= 2(x  3/4)^2 + 9/8  8/8
= 2(x  3/4)^2 + 1/8 
the middle term should of course be (3/2)x in two of the above lines.
I am pretty sure you know where the typo error is. 
r u there gagan
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