How do you solve and check for extraneous solutions in #3sqrt(2x+4)=12#?

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Answer 1 · 2015-08-10T10:23:57

#x = 6#

First, isolate the radical on one side of the equation by dividing both sides by ##

#(color(red)(cancel(color(black)(3))) * sqrt(2x+4))/color(red)(cancel(color(black)(3))) = 12/3#

#sqrt(2x+4) = 4#

Square both sides of the equation to get rid of the square root

#(sqrt(2x+4))^2= 4^2#

#2x+4 = 16#

Finally, isolate #x# on one side of the equation by adding #-4# to both sides and dividing everything by #2#

#2x + color(red)(cancel(color(black)(4))) - color(red)(cancel(color(black)(4))) = 16 - 4#

#(color(red)(cancel(color(black)(2)))x)/color(red)(cancel(color(black)(2))) = 12/2#

#x = color(green)(6)#