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- Oct 14th 2009, 05:26 PM #1

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- Oct 2009
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- Oct 14th 2009, 06:16 PM #2

- Oct 14th 2009, 10:30 PM #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