1. ## Stuck

The instructions say just to solve.
x^2 +2x -4
x^2 + 2x = 4
Now I add (b/2a)squared which would be (4/4)which equals 1
x^2 + 2x +1 = 4 + 1
x^2 +2x + 1= 5
stuck as to what to do next?

2. Hi

x² + 2x + 1 = (x+1)²

3. sorry I don't think you understood because I wrote in wrong
x^2 +2x - 4 = 0
x^2 +2x = 4
then I add 1/2 of the coefficient of x and square it, then add the results to both sides of the equation. Therefore I have (2/2)squared which gives me (4/4) which is 1
x^2 + 2x + 1 = 4 + 1
x^2 + 2x + 1 = 5
x^2 + 2x + 1 - 5 = 5-5
x^2 + 2x - 4 = 0
Now I am stuck if I have done everything else right

4. You did all the hard work to get:
$x^2 + 2x + 1 = 5$

and then undid it all by subtracting 5 from it again.

take a look at the last answer you got:
$x^2 + 2x + 1 = (x+1)^2$

That's why this technique is called "completing the square".

5. Originally Posted by Brama Bull
sorry I don't think you understood because I wrote in wrong
x^2 +2x - 4 = 0
x^2 +2x = 4
then I add 1/2 of the coefficient of x and square it, then add the results to both sides of the equation. Therefore I have (2/2)squared which gives me (4/4) which is 1
x^2 + 2x + 1 = 4 + 1
x^2 + 2x + 1 = 5
This is where you stopped before and this is where you should have stopped! Do you understand why "add 1/2 of the coefficient of x and square it"? The whole purpose is to make one side a "perfect square". That was why running gag said " $x^2+ 2x+ 1= (x+1)^2$".

Can you solve $(x+1)^2= 5$?

x^2 + 2x + 1 - 5 = 5-5
x^2 + 2x - 4 = 0
Now I am stuck if I have done everything else right
These last two lines just undo what you did in the first two lines!