Question

3x2 -6x - 4 =0 how to complete the square?

Answer 1

See below...

Explanation:

We have the quadratic `3x^2-6x-4=0`

First of all, we take out a factor of 3. Don't take it out from the constant however as it can lead to some unnecessary fraction work.

`3x^2-6x-4 => 3[x^2-2x]-4`

Now we write out our initial bracket. To do this we have

`(x+b/2)^2` `=>` in this case `b` is `-2`. Note that we don't include an `x` after the `b`...
Once we have our initial bracket, we subtract the square of `b/2`

`therefore 3[x^2-2x]-4 => 3[(x-1)^2 -1]-4`

Now we must remove the square brackets by multiplying whats in it by the factor on the outside, in this case `3`.

`therefore` we get `3(x-1)^2 -3-4 = 3(x-1)^2 -7`

Final answer `3(x-1)^2 - 7 =0`