complete the square for the equation X^2+2x-3=0

Question

anonymous · Answer

$$x ^{2} + 2x -3$$

anonymous · Answer

$$= x ^{2} +3x -x - 3$$

larseighner · Answer

Don't complete the square and you can factor.

anonymous · Answer

$$= x(x+3) - 1(x+3)$$

anonymous · Answer

$$= (x-1)(x+3)$$

larseighner · Answer

yes

larseighner · Answer

This is a silly exercise, because completing the square is unnecessary to solve it, since the factors are clearly evident on inspection.

anonymous · Answer

i dont get it :/

anonymous · Answer

you don't?
where do you need help? tell me

larseighner · Answer

You can complete the square by adding 4 to both sides

That gives you 

x^2+2x-3 +4 = +4

x^2+2x + 1 = +4

larseighner · Answer

You should recognize that both sides are perfect squares now, to make it clear:
x^2+2x + 1 = +4
(x + 1)(x + 1) = 4

There are several ways to go here.  One is to take the root of both sides:

$$ \large  \sqrt{(x+1)^2} = \pm 2 $$

$$ \large  x+1 = \pm 2 $$

That gives you x+3 and x-1

But this is useless busy work because you could have factored the original expression to begin with.

larseighner · Answer

Another way:

x^2+2x-3 +4 = +4

x^2+2x+1= +4

(x+1)^2 = +4
(x+1)^2  - 4 = 0

This is the difference of two squares, so you should see it factors to

[(x+1)  + 2][(x+1)  - 2] = 0
(x+3)(x-1) = 0

larseighner · Answer

@Maravega33  Are you still with us?  Does one of these explanations make more sense to you?

anonymous · Answer

Use the formula for CTS if you get lost.