How do you complete the square for #x^2 -3x#?

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Answer 1 · 2015-05-24T13:55:22

#x^2-3x = (x-3/2)^2 - 9/4#

To see this, let's multiply out...

#(x-3/2)^2 = (x-3/2)(x-3/2)#

#= x(x-3/2)-3/2(x-3/2)#

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

Subtract #9/4# from both sides to get #x^2-3x = (x-3/2)^2 - 9/4#

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

In your case we have #a=1#, #b=-3# and #c=0#.