How do you solve and check for extraneous solutions in #sqrt(x+2)=x-4#?

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Answer 1 · 2015-08-01T17:07:30

#color(red)(x = 7)# is a solution.
#color(red)(x=2)# is an extraneous solution.

SOLVE:

#sqrt(x+2) = x-4#

Square each side.

#x+2=(x-4)^2#

Remove parentheses.

#x+2 = x^2-8x +16#

Move all terms to the right hand side.

#0=x^2-8x+16-x-2#

Combine like terms.

#0=x^2-9x+14#

Factor:

#0 = (x-7)(x-2)#

#x-7=0# and #x-2=0#

#x = 7# and #x=2#

CHECK FOR EXTRANEOUS SOLUTIONS

#sqrt(x+2) = x-4#

If #x=7#,

#sqrt(7+2) = 7-4#

#sqrt9 = 3#

#3=3#

#x=7# is a solution.

If #x=2#,

#sqrt(2+2) = 2-4#

#sqrt4=-2#

#2 = -2#

This is impossible, so #x=2# is an extraneous solution.