Posted by **gagan** on Thursday, March 26, 2009 at 8:37pm.

how to complete the square for

-2x^2+3x-1?

can you plz help me, and any help is greatly appricieted.

thnx a lot for your effort.

- math -
**Reiny**, Thursday, March 26, 2009 at 9:30pm
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+3x-1

= -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**

- small typo math -
**Reiny**, Thursday, March 26, 2009 at 9:32pm
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.

- math -
**misti**, Thursday, March 26, 2009 at 11:56pm
r u there gagan

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