How do you solve #(x - 5) ^ { 2} - 2( x - 5) - 15= 0#?

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Answer 1 · 2017-10-20T20:14:47

#x=10# or #x=2#

expand brackets:

#(x-5)^2= x^2-10x+25#

#-2(x-5)=-2x+10#

#x^2-10x+25-2x+10-15=0#

collect like terms:

#x^2-12x+20=0#

factorise:

#-10-2 = -12#
#-10*-2 = 20#

#(x-10)(x-2)=0#

therefore, either #x-10# or #x-2# must be #0#.

#x-10=0->x=10#

#x-2=0->x=2#

#x=10# or #x=2#