How do you solve #(x+2)^2-(x-2)^2=(x-1)^2-(x-3)^2#?

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Answer 1 · 2017-06-05T06:17:38

#x=-2#

#(x+2)^2-(x-2)^2=(x-1)^2-(x-3)^2#

#hArr(x+2+x-2)(x+2-x+2)=(x-1+x-3)(x-1-x+3)#

or #(x+cancel2+x-cancel2)(cancelx+2-cancelx+2)=(x-1+x-3)(cancelx-1-cancelx+3)#

or #2x xx 4=(2x-4)xx2#

or #8x=4x-8#

or #8x-4x=-8#

or #4x=-8#

or #x=-2#