How do you solve #sqrt(x^2+5)=3-x# and find any extraneous solutions?
2 Answers
Explanation:
Given:
#sqrt(x^2+5) = 3-x#
Square both sides (noting that this may introduce spurious solutions) to get:
#x^2+5 = 9-6x+x^2#
Subtract
#5 = 9-6x#
Subtract
#-4 = -6x#
Divide both sides by
#2/3 = x#
This is a valid solution of the original equation:
#sqrt((2/3)^2+5) = sqrt(4/9+5) = sqrt(49/9) = 7/3 = (9-2)/3 = 3-2/3#