How do you solve #x^2-5x-3/4=0# by completing the square?

1 Answer
May 13, 2016

Answer:

#x = 5/2+-sqrt(7)#

Explanation:

Premultiply by #4# first to cut down on the fractions involved:

#0 = 4(x^2-5x-3/4)#

#=4x^2-20x-3#

#=(2x)^2-2(2x)(5)-3#

#=(2x-5)^2-5^2-3#

#=(2x-5)^2-28#

#=(2x-5)^2-2^2*7#

#=(2x-5)^2-(2sqrt(7))^2#

#=((2x-5)-2sqrt(7))((2x-5)+2sqrt(7))#

#=(2x-5-2sqrt(7))(2x-5+2sqrt(7))#

#=4(x-5/2-sqrt(7))(x-5/2+sqrt(7))#

So #x = 5/2+-sqrt(7)#