Question

Complete the square -3x^2+12x-7

Answer 1

`-3(x-2)^2+5`

Explanation:

First we can factor a -3 from the
`x^2` term and the
`x` term to get
`-3(x^2-4x)-7`.

Then we want the stuff in the parentheses to have the form of
`(x+b)^2`, or equivalently,
`(x^2+2b+b^2)` . So we can let
`2b = -4`. By solving it, we get
`b = -4/2 = -2`. Then our
`b^2` term should be
`b^2 = (-2)^2 = 4`.

In order to make our
`b^2` term appear in the parentheses, we need to add and subtract our
`b^2` term, so we get
`-3(x^2 - 4x + 4 - 4) -7`.

What we to keep inside our parentheses is
`(x^2 - 4x + 4)` , so we can factor the
`-4` out of parentheses to get
`-3(x^2-4x+4-4)-7 = -3(x^2-4x+4)+(-3)(-4) - 7 = -3(x^2 - 4x + 4) + 12 - 7 = -3(x^2 - 4x + 4) + 5`

Finally, plugging
`b = -2` that we computed earlier into the equation
`(x^2+2b+b^2) = (x+b)^2`, we get
`(x^2 - 4x + 4) = (x-2)^2`.

So we have
`-3(x^2-4x+4)+5 = -3(x-2)^2+5`.

In summary, the procedure is

`-3x^2+12x-7 \\= &-3(x^2-4x) -7 \\= &-3(x^2-4x+4-4)-7 \\= -3(x^2-4x+4) + (-3)(-4)-7\\=-3(x^2-4x+4)+12-7\\= -3(x^2-4x+4)+5 \\= -3(x-2)^2+5`