How do you solve #x^2-6x-9=0# by completing the square?

2 Answers
Jan 19, 2018

Answer:

#(x-3)^2-18#

Explanation:

#x^2-6x-9 = (x^2-6x+9)-18#

#x^2-6x+9 = (x-3)(x-3) = (x-3)^2#

#(x-3)^2-18 = (x^2-6x+9)-(9+9) = x^2-6x+9-9-9#

#= x^2-6x-9#

Jan 19, 2018

Answer:

#x = 3+-3sqrt(2)#

Explanation:

The difference of squares identity can be written:

#A^2-B^2 = (A-B)(A+B)#

Use this with #A=(x-3)# and #B=3sqrt(2)# as follows:

#0 = x^2-6x-9#

#color(white)(0) = x^2-2(x)(3)+3^2-18#

#color(white)(0) = (x-3)^2-(3sqrt(2))^2#

#color(white)(0) = ((x-3)-3sqrt(2))((x-3)+3sqrt(2))#

#color(white)(0) = (x-3-3sqrt(2))(x-3+3sqrt(2))#

Hence:

#x = 3+-3sqrt(2)#