How do you solve #2(x - 1)^2 - 8(x - 1)^2= 0#?

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Answer 1 · 2015-05-21T07:53:02

#(x-1)^2# is common to both the terms of the expression on the left hand side

We can write the equation as #(x-1)^2*(2-8) = 0#

#->-6(x-1)^2 = 0#

Dividing both sides by #-6# we get:

#->(cancel(-6)(x-1)^2)/cancel(-6) = 0/-6#

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

#->x-1=0#

#-> color(green)(x = 1#