How do you solve #x² - 2x = 15# by completing the square?

3 Answers
Jun 10, 2018

Answer:

#x=5# or #x=-3#

Explanation:

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#

Jun 10, 2018

Answer:

#x=5# and #x=-3#

Explanation:

#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#

Jun 10, 2018

Answer:

#x=-3, x=5#

Explanation:

#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#