How do you solve by completing the square $3 x^{2} + 6 x + 12 = 0$ $3x^2 + 6x +12 = 0$ ?

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How do you solve by completing the square $3 x^{2} + 6 x + 12 = 0$ $3x^2 + 6x +12 = 0$ ?

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Answer 1 · 2015-03-30T19:19:44

First divide both sides of the equation by the coefficient of $x^{2}$ $x^2$ (in this case $3$ $3$ )
$x^{2} + 2 x + 4 = 0$ $x^2+2x+4=0$

If $x^{2} + 2 x$ $x^2 + 2x$ are the first 2 terms of an expression from
${(x + a)}^{2}$ $(x+a)^2$
then $a = 1$ $a=1$

Rewrite the left-side of the expression as
$(x^{2} + 2 x + 1) + (3) = 0$ $(x^2+2x+1) + (3) = 0$

${(x + 1)}^{2} = - 3$ $(x+1)^2 = -3$

This can not be solved with any real number value for $x$ $x$
as can be seen from the graph of the original equation below (it never is equal to $0$ $0$ )
graph{3x^2+6x+4 [-16.03, 16, -1.31, 14.71]}

However, if we allow complex solutions
$x + 1 = \pm \sqrt{- 3}$ $x+1 = +-sqrt(-3)$

$x = - 1 - \sqrt{3} i$ $x = -1 -sqrt(3)i$ and $x = - 1 + \sqrt{3} i$ $x=-1+sqrt(3)i$

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How do you solve by completing the square $3 x^{2} + 6 x + 12 = 0$ $3x^2 + 6x +12 = 0$ ?

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How do you solve by completing the square $3 x^{2} + 6 x + 12 = 0$ $3x^2 + 6x +12 = 0$ ?