x^2 + x + 1 = 0
i know the answer and is close to mine i did it like this but is wrong!!
x^2 +x = -1
(x + 1/2) ^ 2 = -1
x+1/2 = + or - squareroot of -1 or i
i know the answer and is close to mine i did it like this but is wrong!!
x^2 +x = -1
(x + 1/2) ^ 2 = -1
x+1/2 = + or - squareroot of -1 or i
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You forgot to actually complete the square ... take half middle coef and square it.Add to both sides.
x² + x + ¼ = –1 + ¼
(x + ½)² = –¾
Now when you take the square root of both sides you'll get ...
x + ½ = ± √(–3/4) = ±½√–3 = ±½i√3
So x = –½ ± ½√3 i
x² + x + ¼ = –1 + ¼
(x + ½)² = –¾
Now when you take the square root of both sides you'll get ...
x + ½ = ± √(–3/4) = ±½√–3 = ±½i√3
So x = –½ ± ½√3 i
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if you are solving by completing the square
x^2 + x = -1
x^2 + x + (1/2)^2 = -1 + (1/2)^2
x^2 + x + 1/4 = -1 + 1/4
(x + 1/2)(x + 1/2) = -3/4
(x + 1/2)^2 = -3/4
x + 1/2 = ± √(-3/4)
x + 1/2 = ± i√3 / 2
x = (-1 ± i√3) / 2
quadratically
x^2 + x + 1 = 0
(-1 ± √-3) / 2
x = (-1 ± i√3) / 2
x^2 + x = -1
x^2 + x + (1/2)^2 = -1 + (1/2)^2
x^2 + x + 1/4 = -1 + 1/4
(x + 1/2)(x + 1/2) = -3/4
(x + 1/2)^2 = -3/4
x + 1/2 = ± √(-3/4)
x + 1/2 = ± i√3 / 2
x = (-1 ± i√3) / 2
quadratically
x^2 + x + 1 = 0
(-1 ± √-3) / 2
x = (-1 ± i√3) / 2