How do you solve #x^2-3x-20=0# by completing the square?

1 Answer
Jan 8, 2017

Answer:

#x = 3/2+-sqrt(89)/2#

Explanation:

The difference of squares identity can be written:

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

Use this with #a=2x-3# and #b=sqrt(89)# as follows:

#0 = 4(x^2-3x-20)#

#color(white)(0) = 4x^2-12x-80#

#color(white)(0) = (2x)^2-2(2x)(3)+9-89#

#color(white)(0) = (2x-3)^2-(sqrt(89))^2#

#color(white)(0) = ((2x-3)-sqrt(89))((2x-3)+sqrt(89))#

#color(white)(0) = (2x-3-sqrt(89))(2x-3+sqrt(89))#

Hence:

#x = 1/2(3+-sqrt(89)) = 3/2+-sqrt(89)/2#