1. ## Completing the square

When completing the square of a quadratic equality (aX^2 + bx + c = 0), step 1 - divide b and c by a, step 2 - divide b/a by 2 and square it. Then add the square to both sides. Got it so far. Step 3 - Convert the left hand side into a squared form, hence (x + b/a)^2.

Step 3 loses me. How do you get (x + b/2a)^2 out of x^2 + b/ax + b/2a^2?? I don't see b/ax anywhere in the new square. To me, (x+ b/2a)^2 looks like only x^2 and b/2a^2 are being added. What happens to the b/ax??

2. you might understand this better by looking at an example

$x^2+y^2-8x+2y+8=0$

this needs to complete the square to get an equation of a circle

first group the x-terms, group the y terms and the constant to the right side.
$
(x^2-8x) + (y^2+2y) = - 8$

you can see that this does not represent

$(x-h)^2+(y-k)^2=r$

the procedure to do this is to take half the coefficient of x, square it, and then add and subtract that result to the original expression

$x^2+bx=x^2+bx+(b/2)^2-(b/2)^2=[x+(b/2)]^2 - (b/2)^2$

this yields

$(x^2-8x+16)-16+(y^2+2y+1)-1=-8$

from which we obtain
$
(x-4)^2+(y+1)^2 = 9$

3. So, lets say you start with a simple equation for a parabola in standard form,
y=x^2+4x+6 (ax^2+bx+c)
and you want to complete the square. You will need to find the third part of this equation, that would be used to make a perfect square with x^2 +4x.
In order to do this, take your b value, divide it by 2, and square it.
Here, 4x is the b value. So,
4/2= 2.
2^2=4
Therefore, the perfect third number of this perfect square is 4.
The ideal perfect square:
y=x^2+4x+4
However, we cannot just go adding and subtracting to equations as we see fit. We get around this by adding zero.
So,
y=x^2+4x+4-4+6
We have not changes this equation, we have merely added zero (4-4=0).
Now, we are free to complete the square.
Do as you would normally do to factor a perfect square. The number you will be using in this case is 2, because when you originally divided 4x by 2, you got 2. It will be positive because 4x is positive. If 4x was negative, the number would be negative. So...
y=(x+2)^2 -4-6
The minus four and minus six are still here because we have not used them in factoring the perfect square. so, the final answer would be
y=(x+2)^2-10
**Note: if the a value in standard form was not one, you would need to factor that out first, and complete the square within the resulting brackets, not forgetting to multiply the c value by the a value when it is removed form the brackets**

4. Okay got it. thanks. one more, could I just look at it this way (x + x-vertex)^2 = c + x-vertex^2 ?