How do you solve and check for extraneous solutions in #sqrt(x+3) =x-3#?

1 Answer
George C.
Jul 31, 2015

Square both sides, solve the resulting quadratic equation and discard the extraneous solution to find: #x=6#

Explanation:

First note that we require #x >= 3# since the left hand side is non-negative.

Now square both sides to get:
#x+3 = x^2-6x+9#

Subtract #x+3# from both sides to get:

#0 = x^2-7x+6 = (x-1)(x-6)#

This has two solutions: #x=1# and #x=6#, but #x=1# is not a solution of the original equation since we require #x >= 3#.

So the only solution to the original equation is #x=6#.

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