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

May 6, 2016

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

Explanation:

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

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

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

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

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

$x + 2 = \sqrt{- 25}$ , $x + 2 = - \sqrt{- 25}$

$x + 2 = \sqrt{- 1 \cdot 25}$, $x + 2 = - \sqrt{- 1 \cdot 25}$

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

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

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