solve x^2+2x-3=0 by completing the square

Question

x^2 +2x -3 =0  

 

solve by completing the square and then by quadratic formula

Answer 1

Hi Marie -

The process of completing the square is pretty simple: Start by moving the constant term to the other side of the equals sign by adding 3 to both sides. Then we have this:

 

x² + 2x + ___  = 3 

 

Now divide the coefficient on the x term in half and then square the result. This is what your constant term is in a perfect square trinomial. 

 

In your problem, the coefficient on the x term is 2. Divide it in half and square the result:

 

2/2 = 1, and 1² = 1 << this is what I add to create a perfect square. But if I add it to one side, I have to add it to the other to keep the equation balanced:

 

(x² + 2x + 1) = 3 + 1

 

Now let's change the notation a bit. We created a perfect square on the left hand side, so let's write it as a perfect square: 

 

(x + 1)² = 4

 

Take the square root of both sides 

 

x + 1 = ± 2

 

subtract 1 from both sides

x= + 2 -1 = 1

x = -2 -1  = -3

 

 

Answer 2

Hi Marie!

 

Remember that the general equation for completing the square is x² + bx + (b/2) = (x + (b/2))²

 

In this case, we have bx=2x or b = 2.

 

So
(2/2)² = 1                  Find (b/2)²

x² + 2x  =  3              Rewrite so all terms containing x are on one side

x² + 2x +1=  3 + 1     Complete the square by adding 1 to each side

(x + 1)² = 4                Factor the perfect square trinomial

(x + 1) = ±2                Find square roots

x=(2 – 1) or x=(-2 – 1) 

 

So x = 1 or -3

 

 

Using quadratic formula:

 

x=  [-b +/- (b²-4ac)^1/2]/2a

 

x=  [-2 +/- (4-4(1)(-3))^1/2]/(2*1)

 

x=  [-2 +/- (4+12)^1/2]/2

 

x=  [-2 +/- (16)^1/2]/2

 

x=  [-2 +/- (4)]/2

 

x= [-2 + 4]/2 or x= [-2-4]/2

 

So  x= 1 or x=-3

 

 

Hope this helps!