How do you solve #sqrt(x+1) = x-1# and find any extraneous solutions?

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Answer 1 · 2016-05-21T14:40:41

#x=3#
# x=0#

First, to remove the #sqrt#, square both sides of the equation, giving:

#x+1=(x-1)^2#

Next, expand the equation out.

#x+1=x^2-2x+1#

Simplify the equation combining like terms.

#x^2-3x=0#

#x(x-3)=0#

Now, you can solve for #x#:

#x=0#
#x=3#

However, if you solved it like this:

#x^2-3x=0#

#x^2=3x#

#x=3#

#x=0# would be a missing solution, this would be an extraneous solution.