How do you solve #sqrt(x+7) = x-5# and find any extraneous solutions?

Question

7122 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-22T06:18:46

#x=9#

#sqrt(x+7) = x-5#

#=> (sqrt(x+7))^2 = (x-5)^2#

#=> x+7 = x^2 - 10x + 25#

#=> x^2 - 11x + 18 = 0#

#=> (x-2)(x-9) = 0#

#=> x-2 = 0 or x-9 = 0#

#=> x = 2 or x = 9#

Test for extraneous solutions:

If #x=2#, we have

#sqrt(2+7) = 3#
#2-5 = -3#

so #x=2# is not a solution.

If #x-9#, we have

#sqrt(9+7) = 4#
#9-5 = 4#

so #x=9# is a solution.

Thus the only solution is #x=9#.