How do you solve #x^2-3x-20=0# by completing the square?
1 Answer
Jan 8, 2017
Explanation:
The difference of squares identity can be written:
#a^2-b^2 = (a-b)(a+b)#
Use this with
#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#