How do you solve and check for extraneous solutions in #sqrt(x+1) + 5 = x#?

1 Answer
Alan P.
Aug 1, 2015

The only valid solution is #x=8#
A second candidate solution #x=3# can be eliminated by checking for the validity of the given equation with #x=3# (and noting that it is not valid).

Explanation:

Given #sqrt(x+1) + 5 = x#

Subtract #5# from both sides
#color(white)("XXXX")##sqrt(x+1) = x-5#

Square both sides (possibly generating an extraneous root at this point)
#color(white)("XXXX")##x+1 = x^2-10x+25#

Subtract #(x+1)# from both sides (and flip sides)
#color(white)("XXXX")##x^2-11x+24 = 0#

Factor
#color(white)("XXXX")##(x-3)(x-8) = 0#

#rArr##color(white)("XXXX")##x=3##color(white)("XXXX")#or#color(white)("XXXX")##x=8#

Substituting #3# for #x# in the Left Side of original equation
#color(white)("XXXX")##sqrt(3+1)+5 = 2+3#
#color(white)("XXXX")##color(white)("XXXX")##!=3#
The solution #x=3# is extraneous.

Substituting #8# for #x# in the Left Side of original equation
#color(white)("XXXX")##sqrt(8+1)+5 = 3+5#
#color(white)("XXXX")##color(white)("XXXX")##=8#
#color(white)("XXXX")##color(white)("XXXX")##=x#
The solution #x=8# is valid.

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