# Thread: x2 + 4x + 5

1. ## x2 + 4x + 5

How would I write this quadratic (x2 + 4x + 5) in standard, factored, and vertex forms?

Thanks.

2. Originally Posted by Andrew00830
How would I write this quadratic (x2 + 4x + 5) in standard, factored, and vertex forms?

Thanks.
$x^2 + 4x + 5$ is already in standard form.

To get vertex form, complete the square: $(x + 2)^2 - 4 + 5 = (x + 2)^2 + 1$.

It should be obvious from the vertex form that it cannot be expressed in factorised form using real linear factors.

3. x^2 + 4x + 5= 0, complete the square,

x^2 + 4x + (4/2)^2 = (4/2)^2 - 5

(x + 2)^2 = 4 - 5 = -1

(x + 2) = (+/-) sqrt (-1); where sqrt (-1) = i

x + 2 = (+/-)i

you have 2 roots: x = -2 + i and x = -2 - i

VERTEX FORM? - that is not clear to me because your equation is in the root finding form. If it for graphing purposes, should it be written in this form (x - h)^2 = 4a(y - k) or what? thanks for clarifying . . . . i may mis-interpret what you mean.

may it be in the form: y = x^2 + 4x + 5?

then x^2 + 4x + 4 = y - 5 + 4, the number 4 was added BS to complete the square.

(x + 2)^2 = y - 1

vertex: (h,k) = (-2,1). see graph attached