How do you solve #2+sqrt[x]=sqrt[x+4]# and find any extraneous solutions?

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Answer 1 · 2016-07-04T00:50:10

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

#4 + 4sqrtx + x = x + 4#

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

#4sqrt(x) = 0#

#(4sqrt(x))^2 = 0^2#

#16x = 0#

#x = 0#

Checking in the original equation we find this solution works. The solution set is #{0}#.

Hopefully this helps!