Completing The Square

[Sandstorm]

Ok, I have to do a sheet on Completing the Square (didn't realize we had to hand it in, threw it out - gotta do it fast :P) with quadratic equations.

The way we learned it went something like this:



z^2 + 4z + 96 = 0.

Make a new trinomial so it's a perfect square trinomial.

(z^2 + 4z + (b/2)^2) + 96 = 0.

So we get

(z^2 + 4z + 4) + 96 = 0.

Remove 96

z^2 + 4z + 4 = 96.

Factor

(z + 2)^2 + 4 = 96

Isolate the square

(z + 2) ^2 = 100

Square root it

z + 2 = +/- 10

Solve

z + 2 = 10

Subtract 2

z = 8

z + 2 = -10

Subtract 2

z = -12

Which is all well and good.

The book shows it done like this:



Skip directly from z^2 + 4z + 96 = 0 to z^2 + 4z = 96

z^2 + 4z = 96

z^2 + 4z + (b/2)^2 = 96 + (b/2)^2

z^2 + 4z + 4 = 96 + 4

This is where my question comes up. In the next step, where the hell did the bolded + 4 go?!

(z + 2)^2 = 100

z + 2 = +/-10

Solve in the same way I learned

z = 8 and z = -12


This second method the book shows seems nice, as it seems shorter. My problem is, where did that +4 go? How did they go about just... removing it?

[TomTuff]

Is there anyway you can take a picture and upload it to imageshack or something? it's difficult to read with the ^ notation.

[Richard]

z^2 + 4z + 4 = (z + 2)^2

They just factorised it. So you end up getting z^2 (z*z), then 2 lots of 2z (2*z), then 4 ( 2*2)

[Sandstorm]

Oh wow, I fail. I read (z + 2) ^ 2 = z^2 + 4z xD. Yeah, ok, I get it. Need to brush up on my FOIL apparently.