How do you solve #sqrt(x+4) = sqrt(3x)# and find any extraneous solutions?

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Answer 1 · 2016-06-12T22:09:08

#sqrt(x + 4) = sqrt(3x)#

#(sqrt(x + 4))^2 = (sqrt(3x))^2#

#x + 4 = 3x#

#4 = 3x - x#

#4 = 2x#

#2 = x#

Checking back in the original equation, we have that #sqrt(6) = sqrt(6)#, so the solution works.

Our solution set is #{x = 2}#.

Hopefully this helps!