
the idea behind the completion of the square is simply using a perfect square trinomial, hmmm usually we do that by using our very good friend Mr Zero, 0.
if we look at the 2nd step, we have a group as x² - x, hmmm so we need a third element, which will be squared.
keeping in mind that the middle term of the perfect square trinomial is simply the product of the roots of "a" and "b", so in this case the middle term is "-x", and the 1st term is x², so we can say that

so that means that our missing third term for a perfect square trinomial is simply 1/2, now we'll go to our good friend Mr Zero, if we add (1/2)², we have to also subtract (1/2)², because all we're really doing is borrowing from Zero, so we'll be including then +(1/2)² and -(1/2)², keeping in mind that 1/4 - 1/4 = 0, so let's do that.
![-3~~ = ~~-2\left[ x^2-x+\left( \cfrac{1}{2} \right)^2 ~~ - ~~\left( \cfrac{1}{2} \right)^2\right]\implies -3=-2\left(x^2-x+\cfrac{1}{4}-\cfrac{1}{4} \right) \\\\\\ -3=-2\left(x^2-x+\cfrac{1}{4} \right)+(-2)-\cfrac{1}{4}\implies -3=-2\left(x^2-x+\cfrac{1}{4} \right)+\cfrac{1}{2} \\\\\\ -3-\cfrac{1}{2}=-2\left(x^2-x+\cfrac{1}{4} \right)\implies -\cfrac{7}{2}=-2\left(x-\cfrac{1}{2} \right)^2\implies \cfrac{7}{4}=\left(x-\cfrac{1}{2} \right)^2](https://tex.z-dn.net/?f=-3~~%20%3D%20~~-2%5Cleft%5B%20x%5E2-x%2B%5Cleft%28%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%20~~%20-%20~~%5Cleft%28%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%5Cright%5D%5Cimplies%20-3%3D-2%5Cleft%28x%5E2-x%2B%5Ccfrac%7B1%7D%7B4%7D-%5Ccfrac%7B1%7D%7B4%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20-3%3D-2%5Cleft%28x%5E2-x%2B%5Ccfrac%7B1%7D%7B4%7D%20%5Cright%29%2B%28-2%29-%5Ccfrac%7B1%7D%7B4%7D%5Cimplies%20-3%3D-2%5Cleft%28x%5E2-x%2B%5Ccfrac%7B1%7D%7B4%7D%20%5Cright%29%2B%5Ccfrac%7B1%7D%7B2%7D%20%5C%5C%5C%5C%5C%5C%20-3-%5Ccfrac%7B1%7D%7B2%7D%3D-2%5Cleft%28x%5E2-x%2B%5Ccfrac%7B1%7D%7B4%7D%20%5Cright%29%5Cimplies%20-%5Ccfrac%7B7%7D%7B2%7D%3D-2%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%5Cimplies%20%5Ccfrac%7B7%7D%7B4%7D%3D%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2)

Hey!
We can use long division to solve this,
Let's write it.

Then, divide

by

to get

.

Multiply the quotient by the divisor...

Subtract

from

.

As you can see, our answer using long division would be 3 with a remainder of 4.
Thanks!
-TetraFish
2x_45 I’m trying to figure it out but if I can get my text from work then I can go back to my dads house tonight and then I get off
When you multiply the equation out and you simplify like terms you get.
2x^2 + 6xy + 4y^2
For AC:
AC= root((a-0)^2 + (b-0)^2)
AC= root((a)^2 + (b)^2)
For BD:
AC= root((a-0)^2 + (0-b)^2)
AC= root((a)^2 + (-b)^2)
AC= root((a)^2 + (b)^2)
Therefore,
AC=BD