How do you solve and check for extraneous solutions in #sqrt(x^2+5)=3-x#?

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Answer 1 · 2015-08-01T16:37:56

#color(red)(x = 2/3)#
There are #color(red)("no")# extraneous solutions.

#sqrt(x^2+5) = 3-x#

Square each side.

#x^2+5=(3-x)^2#

Remove parentheses.

#x^2+5=9-6x+x^2#

Move all terms in #x# the left hand side.

#x^2+5+6x-x^2 =9#

#5+6x =9#

#6x = 9-5 = 4#

#x = 4/6#

#x = 2/3#

There are no extraneous solutions.

Check:

#sqrt(x^2+5) = 3-x#

#sqrt((2/3)^2+5) = 3-2/3#

#sqrt(4/9+5) = (9-2)/3#

#sqrt((45+4)/9) = 7/3#

#sqrt(49/9) = 7/3#

#7/3 = 7/3#