Sandstorm
11-20-2010, 10:54 PM
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?
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?