How do you solve using the completing the square method #x^2 = 2x + 4#?

1 Answer
Oct 26, 2017

Answer:

#x = 3.236" "or" "x=-1.236#

Explanation:

In the equation #x^2 -2x-4=0#, #" "-4# is not #(b/2)^2#

#x^2 -2x " "=4" "larr#move #-4# to the right side

Add #color(blue)((b/2)^2)# to both sides #color(blue)( ((-2)/2)^2 =1)#

#x^2 -2x color(blue)(+1)=4color(blue)(+1)#

#" "(x-1)^2 =5#

#" "x-1 = +-sqrt5#

#" "x = +sqrt5+1color(white)(xxxxxx) or x = -sqrt5+1#

#" "x = +sqrt5+1color(white)(xxxxxx) or x = -sqrt5+1#

#x = 3.236" "or" "x=-1.236#