How do you solve 8+4x-x^2=0 by completing the square?

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Answer 1 · 2018-04-29T13:28:07

#8 = x^2 - 4x = (x^2 -4x + 4) - 4# or #12=(x-2)^2# or

#x = 2 \pm \sqrt{12] = 2 \pm 2 sqrt{3}.#

#8+4x-x^2=0#

#8 = x^2 - 4x#

#8 + (-4/2)^2 = x^2 - 4x + (-4/2)^2 #

# 12 = x^2 -4x + 4#

#12 = (x-2)^2 #

#x - 2 = \pm \sqrt{12} #

#x = 2 \pm sqrt{3}#