How do you solve #sqrt(8 x + 1) = 5# and find any extraneous solutions?

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Answer 1 · 2016-07-07T15:53:35

#x = 3#

Square both sides

#(sqrt(8x+1))^2 = 5^2#

#8x + 1 = 25#

Isolate the variable term

#8x = 24#

Isolate the variable (get rid of the coefficient)

#x = 3#

All values leading up to but not including #-1/8# are when the equation is undefined.