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

Question

6661 views around the world

You can reuse this answer
Creative Commons License

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#