How do you solve $x^2+4x+1=0$ by completing the square?

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Answer 1 · 2016-05-06T16:24:16

$x=-2+-sqrt(3)$

In addition to completing the square, use the difference of squares identity:

$a^2-b^2 = (a-b)(a+b)$

with $a=(x+2)$ and $b=sqrt(3)$ as follows:

$0 = x^2 + 4x + 1$

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

$=(x+2)^2-3$

$=(x+2)^2-(sqrt(3))^2$

$=((x+2)-sqrt(3))((x+2)+sqrt(3))$

$=(x+2-sqrt(3))(x+2+sqrt(3))$

Hence:

$x = -2+-sqrt(3)$

How do you solve x^2+4x+1=0x^2+4x+1=0 by completing the square?