How do you solve #x = 3 +sqrt(x^2 - 9)#?

1 Answer
Aug 5, 2015

Answer:

Isolate the radical; square; use normal solution techniques; then check for extraneous results.
Giving: #x=3#

Explanation:

Given: #x =3+sqrt(x^2-9)#

Isolate radical by subtracting 3 from both sides:
#color(white)("XXXX")##x-3 = sqrt(x^2-9)#
Square both sides
#color(white)("XXXX")##(x-3)^2 = x-9#
Factor (both sides)
#color(white)("XXXX")##(x-3)(x-3) = (x+3)(x-3)#
Note potential solution #(x-3 = 0 rarr x=3)#
Divide both sides by #(x-3)# assuming #x!=3#
#color(white)("XXXX")##color(red)(cancel(color(black)(x)))-3 = color(red)(cancel(color(black)(x)))+3#

#color(white)("XXXX")##-3 = +3#

So the only possible solution we have found is #x=3#

Checking to ensure that this is not an extraneous root introduced by squaring; replace #x# with #3# in the original equation:
#color(white)("XXXX")##(3) ?=? 3 +sqrt((3)^2-9)#?
#color(white)("XXXX")#Yes; equality holds
#color(white)("XXXX")##x=3# is a valid solution