How do you solve $x^2+4x+3=0$ by completing the square?

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Answer 1 · 2016-03-28T00:05:02

$x = -3$ or $x = -1$

Add $1$ to both sides of the equation to get:

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

The left hand side is now a perfect square:

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

So we have:

$(x+2)^2 = 1$

Hence:

$x+2 = +-sqrt(1) = +-1$

Subtract $2$ from both ends to get:

$x = -2+-1$

That is:

$x = -3$ or $x = -1$

How do you solve x^2+4x+3=0x^2+4x+3=0 by completing the square?