How do you solve #sqrt(x+9)-5=x+4# and find any extraneous solutions?

1 Answer
Jun 26, 2016

Answer:

I found:
#x_1=-9#
#x_2=-8#
and no extraneous solutions.

Explanation:

We can write it as:
#sqrt(x+9)=x+4+5#
#sqrt(x+9)=x+9#
square both sides:
#x+9=(x+9)^2#
#x+9=x^2+18x+81#
#x^2+17x+72=0#
use the Quadratic Formula:
#x_(1,2)=(-17+-sqrt(289-288))/2=(-17+-1)/2#
#x_1=-9#
#x_2=-8#
Both works fine into the original equation.