How do you solve by completing the square for #x^2 - 5x = 9#?

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Answer 1 · 2015-06-14T20:06:39

#(x-5/2)^2 = x^2-5x+25/4 = 9+25/4 = 61/4#

Hence: #x = 5/2+-sqrt(61)/2#

#(x-5/2)^2#

#= x^2-5x+25/4#

#= 9+25/4#

#= 36/4+25/4#

#= (36+25)/4#

#= 61/4#

So:

#x - 5/2 = +-sqrt(61/4) = +-sqrt(61)/sqrt(4) = +-sqrt(61)/2#

Add #5/2# to both ends to get:

#x = 5/2+-sqrt(61)/2#

In general,

#ax^2+bx+c = a(x+b/(2a))^2 + (c - b^2/(4a))#