# What's the complete squate of -3x^2+6x-4 ? my answer and the book's answer is different The book's answer is -3(x-1)^2-1 Why is it -1 ? Shouldnt it be -3(x+2)^2 ?? Im confused..

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To complete the square, begin by grouping the `x^2` and `x` terms:

`(-3x^2 +6x)-4`

Now factor out the `-3`

`-3(x^2-2x)-4`

The formula you need now is `(b/2)^2` .

In this case, that is `(-2/2)^2` which equals `1` .

You must both add and subtract it from the problem, so as not to change its value.

The trick is that `1` which is added inside the parentheses is being multiplied by a distributed `-3` . Therefore, you are not really adding `1` , you are subtracting `3` . To counteract that, you must add `3` on the outside.

`-3(x^2-2x+1)-4+3`

Then you must change the part inside the parentheses to a perfect square. Take the square root of the first and last terms, and use the sign from the middle term.

`-3(x-1)^2-1 `

-3x^2 + 6x -4 = -3x^2 + 6x - 3 -1 = -3 (x^2 -2x +1) -1 = -3(x-1)^2 -1 . same as the books answer.

remember, (x-1)^2 = x^2-2x+1