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?

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- Aug 25th 2009, 11:15 AMBrama BullStuck
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? - Aug 25th 2009, 11:29 AMrunning-gag
Hi

x² + 2x + 1 = (x+1)² - Aug 25th 2009, 11:49 AMBrama 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

x^2 + 2x + 1 - 5 = 5-5

x^2 + 2x - 4 = 0

Now I am stuck if I have done everything else right - Aug 25th 2009, 11:58 AMMatt Westwood
You did all the hard work to get:

$\displaystyle x^2 + 2x + 1 = 5$

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

take a look at the last answer you got:

$\displaystyle x^2 + 2x + 1 = (x+1)^2$

That's why this technique is called "completing the square". - Aug 25th 2009, 12:26 PMHallsofIvy
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 "$\displaystyle x^2+ 2x+ 1= (x+1)^2$".

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

Quote:

x^2 + 2x + 1 - 5 = 5-5

x^2 + 2x - 4 = 0

Now I am stuck if I have done everything else right