How do you solve #14= 4+ \sqrt { 17x - 2}#?

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Answer 1 · 2018-03-19T06:47:57

#x= 6 #

#14 = 4 + sqrt(17x-2) #

#14-4 = 4 + sqrt(17x-2) - 4 #

#=> 10 = sqrt(17x-2) #

#=> (10)^2 = (sqrt(17x-2) )^2 #

#=> 100 = 17x-2 #

#=> 102 = 17x #

#=> 102/17 = x #

#x = 6#