How do you solve $x^2+4x+29=0$ by completing the square?

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Answer 1 · 2016-05-06T15:06:50

$color(green)(x = 5i - 2$ , $color(green)(x = -5i -2$

$x^2 + 4x + 29 = 0$

$x^2 + 4x = - 29$

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

$x^2 + 4x + 4 = - 29 + 4$

Using the Identity $color(blue)((a+b)^2 = a^2 + 2ab + b^2$ , we get
$(x+2)^2 = -25$

$x + 2 = sqrt(-25)$ , $x + 2 = -sqrt(-25)$

$x + 2 = sqrt(- 1 * 25)$ , $x + 2 = -sqrt(-1 * 25)$

$x + 2 = i * 5$ , $x + 2 = - i * 5$

$x + 2 = 5i$ , $x + 2 = -5i$

$color(green)(x = 5i - 2$ , $color(green)(x = -5i -2$

How do you solve x^2+4x+29=0x^2+4x+29=0 by completing the square?