How do you solve #x^2= 8x - 15# by completing the square?

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Answer 1 · 2017-07-11T01:46:32

Rearranging first

#x^2-8x+15=0#

#(x-4)^2-4^2+15=0#

#(x-4)^2-1=0#

#(x-4)^2=1#

#(x-4) = +-1#

#x=4+-1#

This gives two values of x, one where you add the 1 and one where you subtract 1

Answer 2 · 2017-07-12T09:31:41

#x =3 or x =5#

#x^2 = 8x -15#

#x^2 -8x = -15" "larr# leave the constant on the right side

Make a perfect square by adding #(b/2)^2# to both sides
#b =-8" "rarr ((-8)/2)^2 =16#

#x^2-8x +16 = -15+16#

#(x-4)^2 = 1#

#x-4 = +-1#

#x -4 = 1" "rarr x =5#

#x-4 = -1" "rarr x = 3#