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

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Answer 1 · 2016-03-28T00:05:02

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

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

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

The left hand side is now a perfect square:

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

So we have:

${(x + 2)}^{2} = 1$ $(x+2)^2 = 1$

Hence:

$x + 2 = \pm \sqrt{1} = \pm 1$ $x+2 = +-sqrt(1) = +-1$

Subtract $2$ $2$ from both ends to get:

$x = - 2 \pm 1$ $x = -2+-1$

That is:

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

How do you solve x2+4x+3=0x^2+4x+3=0 by completing the square?