#1**+3 **

h^{2} + 2h + 5 = 0

First subtract 5 from both sides of the equation.

h^{2} + 2h = 0 - 5

Now we want to add a number to both sides that will cause the left side of the equation to be a perfect square trinomial. So add (2/2)^{2} , which is 1 , to both sides of the equation.

h^{2} + 2h + 1 = 0 - 5 + 1

h^{2} + 2h + 1 = -4

Now that the left side is a perfect square trinomial, it factors like this...

(h + 1)^{2} = -4

And we can add 4 to both sides so the right side = 0 again.

(h + 1)^{2} + 4 = 0

The question didn't say to solve for h , it just said convert to vertex form...I think that is it.

hectictar
Feb 13, 2018

#1**+3 **

Best Answer

h^{2} + 2h + 5 = 0

First subtract 5 from both sides of the equation.

h^{2} + 2h = 0 - 5

Now we want to add a number to both sides that will cause the left side of the equation to be a perfect square trinomial. So add (2/2)^{2} , which is 1 , to both sides of the equation.

h^{2} + 2h + 1 = 0 - 5 + 1

h^{2} + 2h + 1 = -4

Now that the left side is a perfect square trinomial, it factors like this...

(h + 1)^{2} = -4

And we can add 4 to both sides so the right side = 0 again.

(h + 1)^{2} + 4 = 0

The question didn't say to solve for h , it just said convert to vertex form...I think that is it.

hectictar
Feb 13, 2018