How do you solve #x² - 2x = 15# by completing the square?
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Although there is a better method, completing the square method is as such:
rearrange the formula to equate 0:
#x^{2}-2x-15=0#
#(x-1)^2-1-15=0#
Rearrange to get:
#(x-1)^{2}=16#
Square root on both sides to get:
#(x-1) = +-4#
Therefore adding 1 to both sides gives us either: #x=5# or #x=-3#
#x^2-2x-15=0# is in the format of #ax^2+bx+c=0#.
As #c# has already been moved to the left, let's add #(b/2)^2# to both sides and factor:
#x^2-2x=15#
#x^2-2x+(b/2)^2=15+(b/2)^2#
Your value of #b# is the coefficient before the #x# in #bx# (from #ax^2+bx+c=0#):
#x^2-2x+(-2/2)^2=15+(-2/2)^2#
#x^2-2x+(-1)^2=15+(-1)^2#
#x^2-2x+1=15+1#
#(x-1)(x-1)=16#
#(x-1)^2=16#
Now solve for #x#:
#x-1=+-sqrt16#
#x-1=+-4#
#x=4+1# and #x=-4+1#
Therefore, #x=5# and #x=-3#
#x^2-2x-15=0#
#rArr (x-1)^2-1-15#
#rArr (x-1)^2-16=0#
#rArr (x-1)^2=16#
#rArr x-1=pmsqrt16#
#rArr x-1=pm4#
#rArr x=1pm4#
#rArr x=1-4# or #x=1+4#
#rArr x=-3, x=5#
