Question #9e79c

1 Answer
iceman
May 4, 2017

#b=6#

Explanation:

Assuming it is:
#4+sqrt(b-2)=b# => subtract 4 from both sides:
#sqrt(b-2)=b-4# => square both sides:
#b-2=b^2-8b+16# => simplify:
#b^2-9b+18=0# => factor:
#(b-3)(b-6)=0# => solve for b:
#b=3, 6# =>check for extraneous roots: plug in original equation:
#4+sqrt(b-2)=b#
for #b = 3#:
#4+sqrt(3-2)=3#
#4+1!=3# => 3 is an extraneous root and must be rejected
for #b = 6#:
#4+sqrt(6-2)=6#
#4+2=6#
#6=6# => verified #b = 6# is the only valid root.

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