How do you solve #x^2-6x-5=0# by completing the square?

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Answer 1 · 2018-05-31T18:46:58

#x=3+sqrt(14)# or #x=3-sqrt(14)#

Writing
#x^2-2*3x+9-5-9#
This is
#(x-3)^2=14#
#|x-3|=sqrt(14)#
so we get
#x=3+sqrt(14)#
in the other case

#-x+3=sqrt(14)#
#x=3-sqrt(14)#

Answer 2 · 2018-05-31T18:59:31

#x = +-14+3#

#x = +sqrt14 +3 =6.74 or x = -0.74#

#x^2 -6x-5=0#
#x^2 -6x" "=5" "larr# move the constant to the right side.

Complete the square by adding #color(red)((-6/2)^2)# to both sides

#x^2 -6xcolor(red)(+9) =5 color(red)(+9)#

#" "(x-3)^2 = 14" "larr# write as the square of a binomial

#" "x-3 = +-sqrt14" "larr# square root both sides.

#x = +-14+3" "larr# isolate #x#

This gives two solutions for #x:#

#x = +sqrt14 +3 =6.74 or x = -0.74#