How do you solve #sqrt(8-2x) = x-4#?

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Answer 1 · 2017-10-02T20:26:52

#x=2#
#x=4#

#sqrt(8-2x)= x-4#

Square both sides to remove radical:

#8-2x= (x-4)^2#

Expand right hand side:

#8-2x= x^2-8x+16#

Collect like terms on one side of the equation:

#x^2 -6x +8=0#

Factor #x^2 -6x +8=0#:

#(x -2)(x-4)=0=> x= 2# and #x=4#