How do you solve $x^2 + 4x - 12= 0$ using completing the square?

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Answer 1 · 2015-06-11T22:19:31

$0 = x^2+4x-12 = (x+2)^2-4-12 = (x+2)^2-16$
$= (x+2)^2 - 4^2$

Hence $x = -2 +- 4$

That is $x = -6$ or $x = 2$

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

So:

$0 = x^2+4x-12 = x^2+4x+4-4-12$

$=(x+2)^2-4-12 = (x+2)^2-16 = (x+2)^2-4^2$

Add $4^2$ to both ends to get:

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

Hence:

$x+2 = +-4$

Subtract $2$ from both sides to get:

$x = -2+-4$

How do you solve x^2 + 4x - 12= 0x^2 + 4x - 12= 0 using completing the square?