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

Question

1415 views around the world

You can reuse this answer
Creative Commons License

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!