How do you solve #sqrt(u+6)=u# and check your solution?

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Answer 1 · 2017-07-13T19:48:05

#u=3#

#sqrt(u+6) = u#

#u+6=u^2# #->#square both sides to get rid of radical

#0=u^2 - u - 6# #->#subtract #6# and #u# to bring all terms to one side

#0=(u-3)(u+2)# #-># factor

There are two possibilities:

#0=u-3#
#=>u=3#

#0=u+2#
#=>u=-2#

To check your solutions, substitute them back into the original equation.

#u=3#

#sqrt(u+6)=u#
#sqrt(3+6)=3#
#sqrt(9)=3#
#3=3# #->#works

#u=-2#

#sqrt(u+6)=u#
#sqrt(-2+6)=-2#
#sqrt(4)=-2#
#2=-2# #->#does not work

So, the only solution is #u=3#.