How do you solve the equation #x^2-2/3x-26/9=0# by completing the square?

1 Answer
Jan 8, 2017

Answer:

#x = 1/3+-sqrt(3)#

Explanation:

Given:

#x^2-2/3x-26/9 = 0#

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

Use this with #a=(x-1/3)# and #b=sqrt(3)# as follows:

#0 = x^2-2/3x-26/9#

#color(white)(0) = x^2-2/3x+1/9-3#

#color(white)(0) = (x-1/3)^2-(sqrt(3))^2#

#color(white)(0) = ((x-1/3)-sqrt(3))((x-1/3)+sqrt(3))#

#color(white)(0) = (x-1/3-sqrt(3))(x-1/3+sqrt(3))#

Hence:

#x = 1/3+-sqrt(3)#