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

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Answer 1 · 2016-05-20T11:58:13

The solutions are $x = 2$ $color(green)(x = 2$ , $x = - 6$ $color(green)(x = -6$

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

$x^{2} + 4 x = 12$ $x^2 + 4x = 12$

To write the Left Hand Side as a Perfect Square, we add 4 to both sides:

$x^{2} + 4 x + 4 = 12 + 4$ $x^2 + 4x + 4 = 12 + 4$

$x^{2} + 2 \cdot x \cdot 2 + 2^{2} = 16$ $x^2 + 2 * x * 2 + 2^2 = 16$

Using the Identity ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $color(blue)((a+b)^2 = a^2 + 2ab + b^2$ , we get
${(x + 2)}^{2} = 16$ $(x+2)^2 = 16$

$x + 2 = \sqrt{16}$ $x + 2 = sqrt16$ or $x + 2 = - \sqrt{16}$ $x +2 = -sqrt16$

$x + 2 = 4$ $x + 2 = 4$ or $x + 2 = - 4$ $x +2 = -4$

$x = 4 - 2$ $x = 4 -2$ or $x = - 4 - 2$ $x = -4 -2$

$x = 2$ $color(green)(x = 2$ , $x = - 6$ $color(green)(x = -6$

How do you solve x^2 + 4x - 12 = 0x2+4x−12=0x^2 + 4x - 12 = 0 by completing the square?