How do you solve #4+ \sqrt{6x - 8} = x#?

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Answer 1 · 2016-10-09T14:56:50

#sqrt(6x - 8) = x - 4#

#(sqrt(6x - 8))^2 = (x - 4)^2#

#6x - 8 = x^2 - 8x + 16#

#0 = x^2 - 14x + 24#

#0 = (x - 12)(x - 2)#

#x = 12 and 2#

However, #x = 2# is an extraneous solution. The only actual solution is #x = 12#.

Hopefully this helps!