How do you solve by completing the square: #x^2+3x-4=0#?

1 Answer
bp
Apr 3, 2015

x=1 , -4
Take half of the coefficient of x, that would be #3/2# in this case. Square this number and add it and subtract it as shown below.

#x^2# +3x + #9/4# -#9/4#-4=0

#(x+3/2)^2#= #25/4#. Take square root of both sides to have x+#3/2# = #5/2# or #-5/2#

x+#3/2# = #5/2# would give x= #5/2# -#3/2# = 1 and

X+#3/2# = #-5/2# would give x= #-5/2# #-3/2# = -4

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