How do you solve #sqrt(x+6) = 3 + sqrtx#?

1 Answer
Sep 28, 2015

See the explanation.

Explanation:

If #sqrt(x+6) = 3 + sqrtx#, then

#(sqrt(x+6))^2 = (3 + sqrtx)^2# (but not necessarily vie-versa)

So, #x+6 = 9 + 6sqrtx+x#

Which would require
#-3 = 6sqrtx#

But this is not possible, because #sqrtx# is never negative.

So, the equation has no solutions..